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RD Sharma Solutions Class 12 Mathematics Chapter 11 FBQ

Updated on 27 Jan 2022, 02:54 PM IST

The RD sharma solutions are well researched and tested textbooks for the thorough preparations for CBSE class 12 students. Class 12 RD Sharma chapter 11 exercise FBQ solution, deals with the chapter of Higher-order derivatives which is one of the intricate chaptemrs for the students to follow. The RD Sharma class 12th exercise FBQ comes quite handy with the solutions of your complex problems while solving math chapter 11.RD Sharma Solutions So, it is advised to go through the RD Sharma class 12 solution of Higher-Order derivatives exercise FBQ for clearing out the concepts and have a better understanding about the chapter.

RD Sharma Class 12 Solutions Chapter 11 FBQ Higher Order Derivatives - Other Exercise

Higher Order Derivatives Excercise: FBQ

Higher Order Derivatives Exercise Fill in the blanks Question 1

Answer:$\frac{d^{2} y}{d x^{2}}=\frac{5}{16 t^{6}}$
Hint: Differentiate x & y w.r.t t. Use $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$. Again, differentiate w.r.t x.
Given:$y=t^{10}+1 \& x=t^{8}+1$
Solution:
It is given that: $y=t^{10}+1 \& x=t^{8}+1$
Differentiating y w.r.t t,
$\frac{d y}{d t}=10 t^{9}$ …(i) $\left[\because \frac{d}{d x} p^{n}=n p^{n-1}\right]$
Differentiating x w.r.t t,
$\frac{d x}{d t}=8 t^{7}$ …(ii)
So, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$
$\begin{aligned} &\frac{d y}{d x}=\frac{10 t^{9}}{8 t^{7}} \\ & \end{aligned}$
$\frac{d y}{d x}=\frac{5}{4} t^{2}$ …(iii)
Again, differentiate (iii) w.r.t x:
$\begin{aligned} &\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{5}{4} t^{2}\right) \\ & \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=\left(\frac{5}{4}\right) \frac{d}{d t}\left(t^{2}\right) \cdot \frac{d t}{d x}$
$\begin{aligned} &=\left(\frac{5}{4}\right)(2 t) \cdot \frac{1}{8 t^{7}} \\ & \end{aligned}$ $\left[\because \frac{d x}{d t}=8 t^{7}\right]$
$=\frac{10}{32 t^{6}}$
Thus, $\frac{d^{2} y}{d x^{2}}=\frac{5}{16 t^{6}}$

Higher Order Derivatives Exercise Fill in the blanks Question 2

Answer:$\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}} \sec ^{3} \theta$
Hint: Differentiate x & y w.r.t $\theta$. Use $\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$. Again, differentiate w.r.t x.
Given:$y=b \cos \theta \& \quad x=a \sin \theta$
Solution:
It is given that: $x=a \sin \theta \& y=b \cos \theta$
Differentiating y w.r.t t,
$\frac{d y}{d \theta}=-b \sin \theta$ …(i)
Differentiating y w.r.t t,
$\frac{d x}{d \theta}=a \cos \theta$ …(ii)
So, $\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$
$\frac{d y}{d x}=\frac{-b \sin \theta}{a \cos \theta}$ …(iii)
Again, differentiate (iii) w.r.t x:
$\begin{aligned} &\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{-b \sin \theta}{a \cos \theta}\right) \\ & \end{aligned}$
$\frac{d^{2} y}{d x^{2}}=\frac{d}{d \theta}\left(\frac{-b \sin \theta}{a \cos \theta}\right) \cdot \frac{d \theta}{d x}$
$\begin{aligned} &=\frac{d}{d \theta}\left(\frac{-b}{a} \tan \theta\right) \cdot \frac{1}{a \cos \theta} \\ & \end{aligned}$ [ using (ii) ]
$=\frac{-b}{a} \sec ^{2} \theta \cdot\left(\frac{1}{a \cos \theta}\right) \\$
$=\frac{-b}{a^{2}} \sec ^{3} \theta$
Thus, $\frac{d^{2} y}{d x^{2}}=\frac{-b}{a^{2}} \sec ^{3} \theta$

Higher Order Derivatives Exercise Fill in the blanks Question 3

Answer:$\frac{d^{2} y}{d x^{2}}=e^{x}$
Hint: Differentiate y w.r.t x twice to get $\frac{d^{2} y}{d x^{2}}$.
Given:$y=x+e^{x}$
Solution:
It is given that: $y=x+e^{x}$
$\therefore \frac{d y}{d x}=1+e^{x}$
Again, differentiating w.r.t x:
$\therefore \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(1+e^{x}\right)$
Thus, $\frac{d^{2} y}{d x^{2}}=e^{x}$.

Higher Order Derivatives exercise Fill in the blanks question 4

Answer:$\frac{d^{2} y}{d x^{2}}=e^{-x}$
Hint: Differentiate y w.r.t x twice to get $\frac{d^{2} y}{d x^{2}}$
Given:$y=1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots \ldots$
Solution:
It is given that: $y=1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots \ldots$
$\therefore \frac{d y}{d x}=-1+\frac{2 x}{2 !}-\frac{3 x^{2}}{3 !}+\frac{4 x^{3}}{4 !}-\ldots$
Again, differentiating w.r.t x:
$\begin{aligned} &\therefore \frac{d^{2} y}{d x^{2}}=\frac{2}{2 !}-\frac{3(2) x}{3 !}+\frac{4(3) x^{2}}{4 !}-\ldots \\ & \end{aligned}$
$=\frac{2}{2}-\frac{6 x}{6}+\frac{12 x^{2}}{24}-\ldots \\$
$=1-x+\frac{x^{2}}{2 !}-\ldots .$
We know that:
$e^{-x}=1-x+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\ldots$
Thus, $\frac{d^{2} y}{d x^{2}}=e^{-x}$.

Higher Order Derivatives exercise Fill in the blanks question 5

Answer:$\frac{d^{2} y}{d x^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}}$
Hint: Differentiate given equation w.r.t x. Use $\frac{d x}{d y}=\frac{1}{d y / d x}$
Given:$y=x+e^{x}$
Solution:
It is given that: $y=x+e^{x}$
$\begin{aligned} &\therefore \frac{d y}{d x}=1+e^{x} \\ & \end{aligned}$
$\frac{d x}{d y}=\frac{1}{d y / d x} \\$
$\therefore \frac{d x}{d y}=\frac{1}{1+e^{x}}$ …(i)
Again, differentiating w.r.t x:
$\begin{aligned} &\frac{d}{d y}\left(\frac{d x}{d y}\right)=\frac{d}{d y}\left(\frac{1}{1+e^{x}}\right) \\ & \end{aligned}$
$\frac{d^{2} x}{d y^{2}}=\frac{d}{d x}\left(\frac{1}{1+e^{x}}\right) \frac{d x}{d y} \\$
$=\left(\frac{-1}{\left(1+e^{x}\right)^{2}}\right) \frac{d}{d x}\left(e^{x}\right) \cdot\left(\frac{1}{1+e^{x}}\right)$ [ using (i) ]
$=\left(\frac{-1}{\left(1+e^{x}\right)^{2}}\right)\left(e^{x}\right) \cdot\left(\frac{1}{1+e^{x}}\right)$
Thus, $\frac{d^{2} x}{d y^{2}}=\frac{-e^{x}}{\left(1+e^{x}\right)^{3}}$

Higher Order Derivatives exercise Fill in the blanks question 6

Answer: $\frac{d^{2} y}{d x^{2}}=\frac{-1}{x^{2}}$
Hint: Differentiate y w.r.t x twice to get $\frac{d^{2} y}{d x^{2}}$.
Given:$y=\log _{e} x$
Solution:
It is given that: $y=\log _{e} x$
$\therefore \frac{d y}{d x}=\frac{1}{x}$
Again, differentiating w.r.t x:
$\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{-1}{x^{2}} \\ & \end{aligned}$
Thus, $2 \frac{d^{2} y}{d x^{2}}=\frac{-1}{x^{2}} .$.

The RD Sharma class 12th exercise FBQ is the fill in the blank question type that provides very few questions but the important ones that might be asked in the exams. The RD Sharma class 12th exercise FBQ