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NCERT Solutions for Exercise 5.5 Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Exercise 5.5 Class 12 Maths Chapter 5 - Continuity and Differentiability

Edited By Ramraj Saini | Updated on Dec 03, 2023 05:06 PM IST | #CBSE Class 12th
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CBSE Class 12th  Exam Date : 15 Feb' 2025 - 15 Feb' 2025

NCERT Solutions For Class 12 Maths Chapter 5 Exercise 5.5

NCERT Solutions for Exercise 5.5 Class 12 Maths Chapter 5 Continuity and Differentiability are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. In the previous exercise , you have already learned about the differentiation of logarithmic and exponential functions. In exercise 5.5 Class 12 Maths, you will learn about logarithmic differentiation which is not the same as differentiation of logarithmic functions. This trick of differentiation is used for the differentiation of functions raised to the power of functions or variables.

In NCERT solutions for Class 12 Maths chapter 5 exercise 5.5, you will learn that logarithmic differentiation relies on the property of log and chain rule that you have already learned. If you have a good command of the chain rule of differentiation, you can easily solve these problems even without knowing about the logarithmic differentiation. You can go through these Class 12th Maths chapter 5 exercise 5.5 to get in-depth knowledge of this concept. 12th class Maths exercise 5.5 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Continuity and Differentiability Exercise: 5.5

Question:1 Differentiate the functions w.r.t. x. \cos x . \cos 2x .\cos 3x

Answer:

Given function is
y=\cos x . \cos 2x .\cos 3x
Now, take log on both sides
\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \log y = \log \cos x + \log \cos 2x + \log \cos 3x
Now, differentiation w.r.t. x
\log y=\log (\cos x . \cos 2x .\cos 3x)\\ \frac{d(\log y )}{dx} = \frac{\log \cos x}{dx} + \frac{\log \cos 2x}{dx} + \frac{\log \cos 3x}{dx}\\ \frac{1}{y}.\frac{dy}{dx} = (-\sin x)\frac{1}{\cos x}+(-2\sin 2x)\frac{1}{\cos 2x}+(-3\sin3x).\frac{1}{\cos3x}\\ \frac{1}{y}\frac{dy}{dx} = -(\tan x+\tan 2x+\tan 3x) \ \ \ \ \ \ (\because \frac{\sin x }{\cos x} =\tan x)\\ \frac{dy}{dx}=-y(\tan x+\tan 2x+\tan 3x)\\ \frac{dy}{dx}= -\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)
There, the answer is -\cos x\cos 2x\cos 3x(\tan x+\tan 2x+\tan 3x)

Question:2. Differentiate the functions w.r.t. x.

\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}

Answer:

Given function is
y=\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}
Take log on both the sides
\log y=\frac{1}{2}\log\left ( \frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)} \right )\\ \log y = \frac{1}{2} (\log(x-1)+\log(x-2)-\log(x-3)-\log(x-4)-\log(x-5))\\
Now, differentiation w.r.t. x is
\frac{d(\log y)}{dx} = \frac{1}{2} (\frac{d(\log(x-1))}{dx}+\frac{d(\log(x-2))}{dx}-\frac{d(\log(x-3))}{dx}-\frac{d(\log(x-4))}{dx}-\\\frac{d(\log(x-5))}{dx})
\frac{1}{y}\frac{dy}{dx}=\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx}=y\frac{1}{2}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})\\ \frac{dy}{dx} = \frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})
Therefore, the answer is \frac{1}{2}\sqrt {\frac{(x-1) ( x-2)}{(x-3 )(x-4 ) (x-5)}}(\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-3}-\frac{1}{x-4}-\frac{1}{x-5})

Question:3 Differentiate the functions w.r.t. x. (\log x ) ^{\cos x}

Answer:

Given function is
y=(\log x ) ^{\cos x}
take log on both the sides
\log y=\cos x\log (\log x )
Now, differentiation w.r.t x is
\frac{d(\log y)}{dx} = \frac{d(\cos x\log(\log x))}{dx}\\ \frac{1}{y}.\frac{dy}{dx} = (-\sin x)(\log(\log x)) + \cos x.\frac{1}{\log x}.\frac{1}{x}\\ \frac{dy}{dx}= y( \cos x.\frac{1}{\log x}.\frac{1}{x}-\sin x\log(\log x) )\\ \frac{dy}{dx} = (\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )
Therefore, the answer is (\log x)^{\cos x}( \frac{\cos x}{x\log x}-\sin x\log(\log x) )

Question:4 Differentiate the functions w.r.t. x. x ^x - 2 ^{ \sin x }

Answer:

Given function is
y = x ^x - 2 ^{ \sin x }
Let's take t = x^x
take log on both the sides
\log t=x\log x\\
Now, differentiation w.r.t x is
\log t=x\log x\\ \frac{d(\log t)}{dt}.\frac{dt}{dx} = \frac{d(x\log x)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)\\ \frac{1}{t}.\frac{dt}{dx} = \log x +1\\ \frac{dt}{dx} = t(\log x+1)\\ \frac{dt}{dx}= x^x(\log x+1) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because t = x^x )
Similarly, take k = 2^{\sin x}
Now, take log on both sides and differentiate w.r.t. x
\log k=\sin x\log 2\\ \frac{d(\log k)}{dk}.\frac{dk}{dx} = \frac{d(\sin x\log 2)}{dx} \ \ \ \ \ \ \ (by \ chain \ rule)\\ \frac{1}{k}.\frac{dk}{dx} = \cos x \log 2\\ \frac{dk}{dx} = k(\cos x\log 2)\\ \frac{dk}{dx}= 2^{\sin x}(\cos x\log 2) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because k = 2^{\sin x} )
Now,
\frac{dy}{dx} = \frac{dt}{dx}-\frac{dk}{dx}\\ \frac{dy}{dx} = x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)

Therefore, the answer is x^x(\log x+1 )- 2^{\sin x}(\cos x\log 2)

Question:5 Differentiate the functions w.r.t. x. ( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4

Answer:

Given function is
y=( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4
Take log on both sides
\log y=\log [( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4]\\ \log y = 2\log(x+3)+3\log(x+4)+4\log(x+5)
Now, differentiate w.r.t. x we get,
\frac{1}{y}.\frac{dy}{dx} = 2.\frac{1}{x+3}+3.\frac{1}{x+4}+4.\frac{1}{x+5}\\ \frac{dy}{dx}=y\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )\\ \frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )\\ \frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right )\\ \frac{dy}{dx} = (x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)
Therefore, the answer is (x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)

Question:6 Differentiate the functions w.r.t. x. ( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }

Answer:

Given function is
y = ( x+ \frac{1}{x} ) ^ x + x ^{ 1 + \frac{1}{x} }
Let's take t = ( x+ \frac{1}{x} ) ^ x
Now, take log on both sides
\log t =x \log ( x+ \frac{1}{x} )
Now, differentiate w.r.t. x
we get,
\frac{1}{t}.\frac{dt}{dx}=\log \left ( x+\frac{1}{x} \right )+x(1-\frac{1}{x^2}).\frac{1}{\left ( x+\frac{1}{x} \right )} = \frac{x^2-1}{x^2+1}+\log \left ( x+\frac{1}{x} \right )\\ \frac{dt}{dx} = t(\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))\\ \frac{dt}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))
Similarly, take k = x^{1+\frac{1}{x}}
Now, take log on both sides
\log k = ({1+\frac{1}{x}})\log x
Now, differentiate w.r.t. x
We get,
\frac{1}{k}.\frac{dk}{dx}=\frac{1}{x} \left ( 1+\frac{1}{x} \right )+(-\frac{1}{x^2}).\log x = \frac{x^2+1}{x^2}+\frac{-1}{x^2}.\log x\\ \frac{dk}{dx} = t(\frac{x^2+1}{x^2}+\left (\frac{-1}{x^2} \right )\log x)\\ \frac{dk}{dx} = x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} = \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )

Therefore, the answer is \left ( x+\frac{1}{x} \right )^x (\left (\frac{x^2-1}{x^2+1} \right )+\log \left ( x+\frac{1}{x} \right ))+x^{x+\frac{1}{x}}\left (\frac{x^2+1-\log x}{x^2} \right )

Question:7 Differentiate the functions w.r.t. x. (\log x )^x + x ^{\log x }

Answer:

Given function is
y = (\log x )^x + x ^{\log x }
Let's take t = (\log x)^x
Now, take log on both the sides
\log t = x \log(\log x)
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \log (\log x) + x.\frac{1}{x}.\frac{1}{\log x}= \log (\log x)+\frac{1}{\log x}\\ \frac{dt}{dx}= t.(\log (\log x)+\frac{1}{\log x})\\ \frac{dt}{dx} =(\log x)^x(\log (\log x)) + (\log x)^x.\frac{1}{\log x}=(\log x)^x(\log (\log x))+ (\log x )^{x-1}
Similarly, take k = x^{\log x}
Now, take log on both sides
\log k = \log x \log x = (\log x)^2
Now, differentiate w.r.t. x
We get,
\frac{1}{k}\frac{dk}{dx} =2 (\log x).\frac{1}{x} \\ \frac{dt}{dx}= k.\left ( 2 (\log x).\frac{1}{x} \right )\\ \frac{dt}{dx} = x^{\log x}.\left (2 (\log x).\frac{1}{x} \right ) = 2x^{\log x-1}.\log x
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} =(\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x
Therefore, the answer is (\log x)^x(\log (\log x))+ (\log x )^{x-1}+ 2x^{\log x-1}.\log x

Question:8 Differentiate the functions w.r.t. x. (\sin x )^x + \sin ^{-1} \sqrt x

Answer:

Given function is
(\sin x )^x + \sin ^{-1} \sqrt x
Lets take t = (\sin x)^x
Now, take log on both the sides
\log t = x \log(\sin x)
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \log (\sin x) + x.\cos x.\frac{1}{\sin x}= \log (\sin x)+x.\cot x \ \ \ (\because \frac{\cos x}{\sin x}=\cot x)\\ \frac{dt}{dx}= t.(\log (\sin x)+x.\cot x)\\ \frac{dt}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)
Similarly, take k = \sin^{-1}\sqrt x
Now, differentiate w.r.t. x
We get,
\frac{dk}{dt} = \frac{1}{\sqrt{1-(\sqrt x)^2}}.\frac{1}{2\sqrt x}= \frac{1}{2\sqrt{x-x^2}}\\ \frac{dk}{dt}=\frac{1}{2\sqrt{x-x^2}}\\
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} =(\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}
Therefore, the answer is (\sin x)^x(\log (\sin x)+x\cot x)+\frac{1}{2\sqrt{x-x^2}}

Question:9 Differentiate the functions w.r.t. x x ^ {\sin x } + ( \sin x )^{\cos x}

Answer:

Given function is
y = x ^ { \sin x } + ( \sin x )^ {\cos x}
Now, take t = x^{\sin x}
Now, take log on both sides
\log t = \sin x \log x
Now, differentiate it w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \cos x \log x+\frac{1}{x}.\sin x\\ \frac{dt}{dx}=t\left ( \cos x \log x+\frac{1}{x}.\sin x \right )\\ \frac{dt}{dx}= x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )
Similarly, take k = (\sin x)^{\cos x}
Now, take log on both the sides
\log k = \cos x \log (\sin x)
Now, differentiate it w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dt} = (-\sin x)(\log (\sin x)) + \cos x.\frac{1}{\sin x}.\cos x=-\sin x\log(\sin x)+\cot x.\cos x\\ \frac{dk}{dt}= k\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )\\ \frac{dk}{dt}=(\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )
Now,
\frac{dy}{dx} = x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )+ (\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )
Therefore, the answer is x^{\sin x}\left ( \cos x \log x+\frac{1}{x}.\sin x \right )+ (\sin x)^{\cos x}\left ( -\sin x\log(\sin x)+\cot x.\cos x \right )

Question:10 Differentiate the functions w.r.t. x. x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }

Answer:

Given function is
x ^ {x \cos x} + \frac{x^2 + 1 }{x^2 -1 }
Take t = x^{x\cos x}
Take log on both the sides
\log t =x\cos x \log x
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} = \cos x\log x-x\sin x\log x + \frac{1}{x}.x.\cos x\\ \frac{dt}{dx}= t.\left (\log x(\cos x-x\sin x)+ \cos x \right ) = x^{x\cos x}\left ( \log x(\cos x-x\sin x)+ \cos x \right )
Similarly,
take k = \frac{x^2+1}{x^2-1}
Now. differentiate it w.r.t. x
we get,
\frac{dk}{dx} = \frac{2x(x^2-1)-2x(x^2+1)}{(x^2-1)^2} = \frac{2x^3-2x-2x^3-2x}{(x^2-1)^2} = \frac{-4x}{(x^2-1)^2}
Now,
\frac{dy}{dx} = \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx} = x^{x\cos x}\left ( \log x(\cos x-x\sin x)+ \cos x \right )-\frac{4x}{(x^2-1)^2}
Therefore, the answer is x^{x\cos x}\left ( \cos x(\log x+1)-x\sin x\log x\right )-\frac{4x}{(x^2-1)^2}

Question:11 Differentiate the functions w.r.t. x. ( x \cos x )^ x + ( x \sin x )^{1/ x}

Answer:

Given function is
f(x)=( x \cos x )^ x + ( x \sin x )^{1/ x}
Let's take t = (x\cos x)^x
Now, take log on both sides
\log t =x\log (x\cos x) = x(\log x+\log \cos x)
Now, differentiate w.r.t. x
we get,
\frac{1}{t}\frac{dt}{dx} =(\log x+\log \cos x)+x(\frac{1}{x}+\frac{1}{\cos x}.(-\sin x))\\ \frac{dt}{dx} = t(\log x + \log \cos x+1-x\tan x) \ \ \ \ \ \ \ \ \ (\because \frac{\sin x}{\cos x}= \tan x)\\ \frac{dt}{dx}= (x\cos x)^x(\log x + \log \cos x+1-x\tan x)\\ \frac{dt}{dx}=(x\cos x)^x(+1-x\tan x+\log (x\cos x))
Similarly, take k = (x\sin x)^{\frac{1}{x}}
Now, take log on both the sides
\log k = \frac{1}{x}(\log x+\log \sin x)
Now, differentiate w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dx} =(\frac{-1}{x^2})(\log x+\log \sin x)+\frac{1}{x}(\frac{1}{x}+\frac{1}{\sin x}.(\cos x))\\ \frac{dk}{dx} = \frac{k}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x}) \ \ \ \ \ \ \ \ \ (\because \frac{\cos x}{\sin x}= \cot x)\\ \frac{dk}{dx}=\frac{(x\sin x)^{\frac{1}{x}}}{x^2}(-\log x - \log \sin x+\frac{1}{x^2}+\frac{\cot x}{x})\\ \frac{dk}{dx}=(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}
Now,
\frac{dy}{dx}= \frac{dt}{dx}+\frac{dk}{dx}
\frac{dy}{dx}= (x\cos x)^x(+1-x\tan x+\log (x\cos x))+(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}
Therefore, the answer is (x\cos x)^x(1-x\tan x+\log (x\cos x))+(x\sin x)^{\frac{1}{x}}\frac{(x\cot x+1-(\log x\sin x))}{x^2}

Question:12 Find dy/dx of the functions given in Exercises 12 to 15

x ^ y + y ^ x = 1.

Answer:

Given function is
f(x)=x ^ y + y ^ x = 1
Now, take t = x^y
take log on both sides
\log t = y\log x
Now, differentiate w.r.t x
we get,
\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})
Similarly, take k = y^x
Now, take log on both sides
\log k = x\log y
Now, differentiate w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})
Now,
f^{'}(x)= \frac{dt}{dx}+\frac{dk}{dx}= 0

( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) + (y^x)(\log y+\frac{x}{y}\frac{dy}{dx}) = 0\\ \frac{dy}{dx}(x^y(\log x)+xy^{x-1}) = -(yx^{y-1}+y^x(\log y))\\ \frac{dy}{dx}= \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}

Therefore, the answer is \frac{ -(yx^{y-1}+y^x(\log y))}{(x^y(\log x)+xy^{x-1})}

Question:13 Find dy/dx of the functions given in Exercises 12 to 15.

y^x = x ^y

Answer:

Given function is
f(x)\Rightarrow x ^ y = y ^ x
Now, take t = x^y
take log on both sides
\log t = y\log x
Now, differentiate w.r.t x
we get,
\frac{1}{t}\frac{dt}{dx} = \frac{dy}{dx}(\log x)+y\frac{1}{x}=\frac{dy}{dx}(\log x)+\frac{y}{x}\\ \frac{dt}{dx}= t(\frac{dy}{dx}(\log x)+\frac{y}{x})\\ \frac{dt}{dx}= ( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x})
Similarly, take k = y^x
Now, take log on both sides
\log k = x\log y
Now, differentiate w.r.t. x
we get,
\frac{1}{k}\frac{dk}{dx} = (\log y)+x\frac{1}{y}\frac{dy}{dx}=\log y+\frac{x}{y}\frac{dy}{dx}\\ \frac{dk}{dx}= k(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dk}{dx}= (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})
Now,
f^{'}(x)\Rightarrow \frac{dt}{dx}= \frac{dk}{dx}

( x^y)(\frac{dy}{dx}(\log x)+\frac{y}{x}) = (y^x)(\log y+\frac{x}{y}\frac{dy}{dx})\\ \frac{dy}{dx}(x^y(\log x)-xy^{x-1}) = (y^x(\log y)-yx^{y-1})\\ \frac{dy}{dx}= \frac{ y^x(\log y)-yx^{y-1}}{(x^y(\log x)-xy^{x-1})} = \frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )

Therefore, the answer is \frac{x}{y}\left ( \frac{y-x\log y}{x-y\log x}\right )

Question:14 Find dy/dx of the functions given in Exercises 12 to 15. ( \cos x )^y = ( \cos y )^x

Answer:

Given function is
f(x)\Rightarrow (\cos x) ^ y = (\cos y) ^ x
Now, take log on both the sides
y\log \cos x = x \log \cos y
Now, differentiate w.r.t x
\frac{dy}{dx}(\log \cos x)-y\tan x = \log \cos y-x\tan y\frac{dy}{dx}
By taking similar terms on the same side
We get,
(\frac{dy}{dx}(\log \cos x)-y\tan x) = (\log \cos y-x\tan y\frac{dy}{dx})\\ \frac{dy}{dx} \left (\log \cos x+(\cos y)^x.x\tan y) \right )= \left ( \log \cos y+(\cos x)^y.y\tan x \right )\\ \frac{dy}{dx}= \frac{\left (\log \cos y+y\tan x \right )}{\left ( \log \cos x+x\tan y) \right )} = \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}

Therefore, the answer is \frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}

Question:15 Find dy/dx of the functions given in Exercises 12 to 15. xy = e ^{x-y}

Answer:

Given function is
f(x)\Rightarrow xy = e ^{x-y}
Now, take log on both the sides
\log x+\ log y = (x-y)(1) \ \ \ \ \ \ \ \ \ \ \ \ (\because \log e = 1)\\ \log x+\ log y = (x-y)
Now, differentiate w.r.t x
\frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}
By taking similar terms on same side
We get,
(\frac{1}{y}+1)\frac{dy}{dx}=1-\frac{1}{x}\\ \frac{y+1}{y}.\frac{dy}{dx}= \frac{x-1}{x}\\ \frac{dy}{dx}= \frac{y}{x}.\frac{x-1}{y+1}
Therefore, the answer is \frac{y}{x}.\frac{x-1}{y+1}

Question:16 Find the derivative of the function given by f (x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8) and hence find

f ' (1)

Answer:

Given function is
y = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)
Take log on both sides
\log y =\log (1 + x) + \log (1 + x^2) +\log (1 + x^4) +\log (1 + x^8)
NOW, differentiate w.r.t. x
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8}\\ \frac{dy}{dx}=y.\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )\\ \frac{dy}{dx}= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )
Therefore, f^{'}(x)= (1 + x) (1 + x^2) (1 + x^4) (1 + x^8).\left ( \frac{1}{1+x}+ \frac{2x}{1+x^2}+ \frac{4x^3}{1+x^4}+ \frac{8x^7}{1+x^8} \right )
Now, the vale of f^{'}(1) is
f^{'}(1)= (1 + 1) (1 + 1^2) (1 + 1^4) (1 + 1^8).\left ( \frac{1}{1+1}+ \frac{2(1)}{1+1^2}+ \frac{4(1)^3}{1+1^4}+ \frac{8(1)^7}{1+1^8} \right )\\ f^{'}(1)=16.\frac{15}{2} = 120

Question:17 (1) Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
(i) by using product rule

Answer:

Given function is
f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)
Now, we need to differentiate using the product rule
f^{'}(x)=\frac{d((x^2 - 5x + 8))}{dx}. (x^3 + 7x + 9)+(x^2 - 5x + 8).\frac{d( (x^3 + 7x + 9))}{dx}\\
= (2x-5).(x^3+7x+9)+(x^2-5x+8)(3x^2+7)\\ =2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56\\ = 5x^4 -20x^3+45x^2-52x+11
Therefore, the answer is 5x^4 -20x^3+45x^2-52x+11

Question:17 (2) Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
(ii) by expanding the product to obtain a single polynomial.

Answer:

Given function is
f(x)=(x^2 - 5x + 8) (x^3 + 7x + 9)
Multiply both to obtain a single higher degree polynomial
f(x) = x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)
= x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72
= x^5-5x^4+15x^3-26x^2+11x+72
Now, differentiate w.r.t. x
we get,
f^{'}(x)=5x^4-20x^3+45x^2-52x+11
Therefore, the answer is 5x^4-20x^3+45x^2-52x+11

Question:17 (3) Differentiate (x^2 - 5x + 8) (x^3 + 7x + 9) in three ways mentioned below:
(iii) by logarithmic differentiation.
Do they all give the same answer?

Answer:

Given function is
y=(x^2 - 5x + 8) (x^3 + 7x + 9)
Now, take log on both the sides
\log y = \log (x^2-5x+8)+\log (x^3+7x+9)
Now, differentiate w.r.t. x
we get,
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{x^2-5x+8}.(2x-5) + \frac{1}{x^3+7x+9}.(3x^2+7)\\ \frac{dy}{dx}= y.\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\ \frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9).\left ( \frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)} \right )\\ \frac{dy}{dx} = (2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)\\ \frac{dy}{dx} = 5x^4-20x^3+45x^2-56x+11
Therefore, the answer is 5x^4-20x^3+45x^2-56x+11
And yes they all give the same answer

Question:18 If u, v and w are functions of x, then show that \frac{d}{dx} ( u,v,w) = \frac{du}{dx} v. w +u . \frac{dv }{dx } v. w+ u . \frac{dv}{dx } . w+u.v \frac{dw}{dx} in two ways - first by repeated application of product rule, second by logarithmic differentiation.

Answer:

It is given that u, v and w are the functions of x
Let y = u.v.w
Now, we differentiate using product rule w.r.t x
First, take y = u.(vw)
Now,
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{d(v.w)}{dx}.u -(i)
Now, again by the product rule
\frac{d(v.w)}{dx}= \frac{dv}{dx}.w + \frac{dw}{dx}.v
Put this in equation (i)
we get,
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)
Hence, by product rule we proved it

Now, by taking the log
Again take y = u.v.w
Now, take log on both sides
\log y = \log u + \log v + \log w
Now, differentiate w.r.t. x
we get,
\frac{1}{y}.\frac{dy}{dx} = \frac{1}{u}.\frac{du}{dx}+\frac{1}{v}\frac{dv}{dx}+\frac{1}{w}.\frac{dw}{dx}\\ \frac{dy}{dx}= y. \left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\ \frac{dy}{dx} = (u.v.w)\left ( \frac{v.w.\frac{du}{dx}+u.w.\frac{dv}{dx}+u.v.\frac{dw}{dx}}{u.v.w} \right )\\
\frac{dy}{dx}= \frac{du}{dx}.(v.w) + \frac{dv}{dx}.(u.w) + \frac{dw}{dx}.(u.v)
Hence, we proved it by taking the log

More About NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5:-

Class 12 Maths ch 5 ex 5.5 consists of questions related to finding differentiation of functions raised to the power of functions. These types of questions can be solved using the concept called logarithmic differentiation. In NCERT book exercise 5.5 Class 12 Maths which you learn this concept through solving problems based on this concept. There are 10 questions from this concept given in the exercise 5.5 class 12 maths. Also, you can solve two examples give before this exercise which will help you to get conceptual clarity. The proof of this concept is also given before this exercise. You can prove the given definition by yourself using the chain rule and logarithmic property.

Also Read| Continuity and Differentiability Class 12th Chapter 5 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.5:-

  • NCERT Solutions for Class 12 Maths chapter 5 exercise 5.5 are beneficial when students are getting difficulty while solving the NCERT syllabus exercise 5.5 Class 12 Maths problems.
  • Class 12 Maths chapter 5 exercise 5.5 solutions are descriptive so you can easily understand the solutions.
  • Class 12th Maths chapter 5 exercise 5.5 solutions can be used for reference when you are solving logarithmic differentiation.
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Key Features Of NCERT Solutions for Exercise 5.5 Class 12 Maths Chapter 5

  • Comprehensive Coverage: The solutions encompass all the topics covered in ex 5.5 class 12, ensuring a thorough understanding of the concepts.
  • Step-by-Step Solutions: In this class 12 maths ex 5.5, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
  • Accuracy and Clarity: Solutions for class 12 ex 5.5 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
  • Conceptual Clarity: In this 12th class maths exercise 5.5 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
  • Inclusive Approach: Solutions for ex 5.5 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
  • Relevance to Curriculum: The solutions for class 12 maths ex 5.5 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

Also see-

NCERT Solutions of Class 12 Subject Wise

Subject Wise NCERT Exampler Solutions

Happy learning!!!

Frequently Asked Questions (FAQs)

1. Does logarithmic differentiation and differentiation of logarithmic function is same ?

No,  logarithmic differentiation and differentiation of logarithmic function are different concepts.

2. What is use of logarithmic differentiation ?

Logarithmic differentiation is useful for differentiating the function raised to the power of some variable or function.

3. What is weightage of Vector Algebra if the CBSE Class 12 Maths board exam ?

The weightage of Vector Algebra is 7 marks in the CBSE Class 12 Maths board exam. For good score follow NCERT book. To solve more problems NCERT exemplar and previous year papers can be used.

4. Can i get CBSE Class 12 Syllabus ?

Click on the link to get CBSE Class 12 Syllabus.

5. What is the application process for CBSE Class 12 ?

Click on the link to get application process for CBSE Class 12

6. What is the exam duration of CBSE Class 12 Maths ?

Total of 3 hours will be given to you to complete the CBSE Class 12 Maths paper.

7. What is the differentiation of e^(2x) ?

The differentiation of e^(2x) is 2 e^(2x).

8. Find the differentiation of 1/x ?

d(1/x)/dx = -1/x^2

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Questions related to CBSE Class 12th

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer
Jefferson 25 September,2024
i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

  1. Re-evaluate Your Study Strategies:

    • Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
    • Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
    • Practice Regularly: Consistent practice is key to mastering chemistry.

  2. Consider Professional Help:

    • Tutoring: A tutor can provide personalized guidance and support.
    • Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.

  3. Explore Alternative Options:

    • Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
    • Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.

  4. Focus on NEET 2025 Preparation:

    • Stay Dedicated: Continue your NEET preparation with renewed determination.
    • Utilize Resources: Make use of study materials, online courses, and mock tests.

  5. Seek Support:

    • Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
    • Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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Kanishka kaushikii 17 September,2024
i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024
i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024
Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024
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Popular CBSE Class 12th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
Read More

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