RD Sharma Class 12 Exercise 10.8 Differentiation Solutions Maths

RD Sharma Class 12 Exercise 10.8 Differentiation Solutions Maths - Download PDF Free Online

Updated on Jan 20, 2022 05:45 PM IST

RD Sharma class 12th exercise 10.8 is the holy grail of many students preparing for board exams. Since maths is a complex subject, the RD Sharma class 12 chapter 10 exercise 10.8 solution is a must-have for students who cannot get their doubts cleared. To score high in exams, students need to practice maths daily and prepare beforehand instead of stressing out the last few months. With the help of RD Sharma solutions, they can practice at home and check their answers to see if they have their concepts cleared.

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RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise
Differentiation Excercise: 10.8
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Students can also use the class 12 RD Sharma chapter 10 exercise 10.8 solution to solve their homework. Teachers tend to give homework questions from RD Sharma solutions so students can answer even complex questions by using the book. If students are thorough with the RD Sharma solutions, they might find some common questions from the book in their board exams.

Also Read - RD Sharma Solution for Class 9 to 12 Maths
The RD Sharma class 12th exercise 10.8 is a top choice for hundreds of students who have excelled in their exams after practicing the book. The 10th chapter of the Class 12 maths book has the topic Differentiations. This part will discuss in detail the basic concept of differentiation of inverse trigonometric functions. It will also teach students about Differentiation of a function with respect to another function, Differentiation by using trigonometric substitutions, Differentiation of implicit functions, and the like. Exercise 10.8 includes 15 on linear trigonometric equations that need to be solved by students.

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

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Differentiation Excercise: 10.8

Differentiation exercise 10.8 question 1

Answer:

2 \sec^{2} (x^{2}) \tan (x^{2})

Hint:

\begin{aligned} Let u = \sec^{2} (x^{2}); v = x^{2} \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} \end{aligned}

Given:

\sec^{2} (x^{2}) w.r.t x^{2}

Explanation:

\begin{aligned} Let u = \sec^{2} (x^{2}), v = x^{2} \\ \frac{d u}{d x} = 2 \sec (x^{2}) [\sec x^{2} \tan x^{2} \times 2 x] \end{aligned}

= 4 x \sec^{2} (x^{2}) \tan (x^{2})

(chain rule)

\begin{aligned} \frac{d v}{d x} = 2 x \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{4 x \sec^{2} (x^{2}) \tan (x^{2})}{2 x} \end{aligned}

= 2 \sec^{2} (x^{2}) \tan (x^{2})

Differentiation exercise 10.8 question 3

Answer:

x (\log x)^{x - 1} {1 + \log x \cdot \log (\log x)}

Hint:

Let u = (\log x)^{x}, v = \log x

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}

Given:

(\log x)^{x} w.r.t \log x

Explanation:

Let u = (\log x)^{x}, v = \log x

u = (\log x)^{x}

Taking log both sides,

\begin{aligned} \log u = \log (\log x)^{x} \\ \log u = x \log (\log x) \end{aligned}

Differentiate both sides w.r.t.

x

\begin{aligned} \frac{1}{u} \frac{d u}{d x} = x [\frac{1}{\log x} \times \frac{1}{x}] + \log (\log x) \times 1 \\ \frac{1}{u} \frac{d u}{d x} = (\log x)^{- 1} + \log (\log x) \\ \frac{d u}{d x} = u [(\log x)^{- 1} + \log (\log x)] \end{aligned}

\begin{aligned} \frac{d u}{d x} = u [\frac{1}{\log x} + \log (\log x)] \\ \frac{d u}{d x} = \frac{u}{\log x} [1 + \log x \log (\log x)] \\ \frac{d u}{d x} = (\log x)^{x - 1} [1 + \log x \cdot \log (\log x)] \end{aligned}

\begin{aligned} v = \log x \\ \frac{d v}{d x} = \frac{1}{x} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{(\log x)^{x - 1} [1 + \log x \cdot \log (\log x)]}{\frac{1}{x}}

= x (\log x)^{x - 1} {1 + \log x \cdot \log (\log x)}

Differentiation exercise 10.8 question 4(i)

Answer: 1
Hint:

u = \sin^{- 1} \sqrt{1 - x^{2}}, v = \cos^{- 1} x

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}

Given:

\sin^{- 1} \sqrt{1 - x^{2}} w.r.t \cos^{- 1} x, x \in (0, 1)

Explanation:

Let u = \sin^{- 1} \sqrt{1 - x^{2}}, v = \cos^{- 1} x

\begin{aligned} Let u = \sin^{- 1} \sqrt{1 - x^{2}}, v = \cos^{- 1} x \\ u = \sin^{- 1} \sqrt{1 - x^{2}} \\ Let x = \cos θ \\ u = \sin^{- 1} \sqrt{1 - \cos^{2} θ} \\ u = \sin^{- 1} (\sin θ) \end{aligned}

\begin{aligned} x \in (0, 1) \\ \cos θ \in (0, 1) θ \in (0, \frac{π}{2}) \\ u = \sin^{- 1} (\sin θ) = θ \end{aligned}

\begin{aligned} u = \cos^{- 1} x \\ \frac{d u}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \\ v = \cos^{- 1} x \end{aligned}

\begin{aligned} \frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{- 1}{\frac{\sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}}}} = 1 \end{aligned}

Differentiation exercise 10.8 question 4(ii)

Answer:

- 1

Hint:

Let u = \sin^{- 1} \sqrt{1 - x^{2}}, v = \cos^{- 1} x

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}

Given:

\sin^{- 1} \sqrt{1 - x^{2}} w.r.t \cos^{- 1} x, x \in (- 1, 0)

Explanation:

Let u = \sin^{- 1} \sqrt{1 - x^{2}}, v = \cos^{- 1} x

\begin{aligned} u = \sin^{- 1} \sqrt{1 - x^{2}} \\ Let x = \cos θ \\ u = \sin^{- 1} \sqrt{1 - \cos^{2} θ} \end{aligned}

\begin{aligned} u = \sin^{- 1} (\sin θ) \\ x \in (- 1, 0) \\ \cos θ \in (- 1, 0) θ \in (- \frac{π}{2}, 0) \end{aligned}

\begin{aligned} u = \sin^{- 1} (\sin θ) = - θ θ \in (- \frac{π}{2}, 0) \\ u = - \cos^{- 1} x \end{aligned}

\begin{aligned} \frac{d u}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \\ v = \cos^{- 1} x \\ \frac{d v}{d x} = \frac{1}{\sqrt{1 - x^{2}}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{- 1}{\sqrt{1 - x^{2}}}}{\frac{1}{\sqrt{1 - x^{2}}}} = - 1

Differentiation exercise 10.8 question 5(i)

Answer:

\frac{- 1}{x}

Hint:

Let u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}), v = \sqrt{1 - 4 x^{2}}

Given:

\sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) w.r.t \sqrt{1 - 4 x^{2}}

Explanation:

Let 2 x = \cos θ

\begin{aligned} u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) \\ u = \sin^{- 1} (2 \cos θ \sqrt{1 - (\cos θ)^{2}}) \end{aligned}

\begin{aligned} = \sin^{- 1} (2 \cos θ \sqrt{1 - \cos^{2} θ}) \\ = \sin^{- 1} (2 \cos θ \sin θ) \\ = \sin^{- 1} (\sin 2 θ) \end{aligned}

Now we try to find range of

θ

x \in (\frac{- 1}{2 \sqrt{2}}, \frac{1}{2 \sqrt{2}})

2 x \in (\frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

\cos θ \in (\frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) [\begin{array}{l} \cos^{- 1} (\frac{1}{\sqrt{2}}) = \frac{π}{4} \\ & \cos^{- 1} (\frac{- 1}{\sqrt{2}}) = \frac{3 π}{4} \end{array}]

\begin{aligned} θ \in (\frac{π}{4}, \frac{3 π}{4}) \\ 2 θ \in (\frac{π}{2}, \frac{3 π}{2}) \end{aligned}

\begin{aligned} u = \sin^{- 1} (\sin 2 θ) \\ = π - 2 θ \\ = π - 2 \cos^{- 1} (2 x) \end{aligned}

[\begin{array}{l} 2 x = \cos θ \\ θ = \cos^{- 1} (2 x) \end{array}]

\frac{d u}{d x} = 0 - 2 [\frac{- 1}{\sqrt{1 - (2 x)^{2}}} \times 2]

= \frac{4}{\sqrt{1 - x^{2}}}

\begin{aligned} v = \sqrt{1 - 4 x^{2}} \\ \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} (- 8 x) \\ = \frac{- 4 x}{\sqrt{1 - x^{2}}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{4}{\sqrt{1 - x^{2}}}}{\frac{- 4 x}{\sqrt{1 - x^{2}}}} = \frac{- 1}{x}

Differentiation exercise 10.8 question 5(ii)

Answer:

\frac{1}{x}

Hint:

: Let u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}), v = \sqrt{1 - 4 x^{2}}

Given:

\sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) w.r.t \sqrt{1 - 4 x^{2}}

Explanation:

Let 2 x = \cos θ

\begin{aligned} u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) \\ u = \sin^{- 1} (2 \cos θ \sqrt{1 - (\cos θ)^{2}}) \end{aligned}

\begin{aligned} = \sin^{- 1} (2 \cos θ \sqrt{1 - \cos^{2} θ}) \\ = \sin^{- 1} (2 \cos θ \sin θ) \\ = \sin^{- 1} (\sin 2 θ) \end{aligned}

Now we try to find the range of

θ

\begin{aligned} x \in (\frac{1}{2 \sqrt{2}}, \frac{1}{2}) \\ 2 x \in (\frac{1}{\sqrt{2}}, 1) \\ \cos θ \in (\frac{1}{\sqrt{2}}, 1) \end{aligned}

\begin{aligned} θ \in (0, \frac{π}{4}) \\ 2 θ \in (0, \frac{π}{2}) \\ u = \sin^{- 1} (\sin 2 θ) \end{aligned}

= 2 θ [\begin{array}{l} 2 x = \cos θ \\ θ = \cos^{- 1} (2 x) \end{array}]

\begin{aligned} u = 2 \cos^{- 1} (2 x) \\ \frac{d u}{d x} = 2 [\frac{- 1}{\sqrt{1 - (2 x)^{2}}} \times 2] \end{aligned}

= \frac{- 4}{\sqrt{1 - 4 x^{2}}}

\begin{aligned} v = \sqrt{1 - 4 x^{2}} \\ \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} (- 8 x) \end{aligned}

= \frac{- 4 x}{\sqrt{1 - 4 x^{2}}}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{- 4}{\sqrt{1 - x^{2}}}}{\frac{- 4 x}{\sqrt{1 - 4 x^{2}}}} = \frac{1}{x}

Differentiation exercise 10.8 question 5(iii)

Answer:

- \frac{1}{x}

Hint:

Let u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}), v = \sqrt{1 - 4 x^{2}}

Given:

\sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) w.r.t \sqrt{1 - 4 x^{2}}

Explanation:

Let 2 x = \cos θ

\begin{aligned} u = \sin^{- 1} (4 x \sqrt{1 - 4 x^{2}}) \\ u = \sin^{- 1} (2 \cos θ \sqrt{1 - (\cos θ)^{2}}) \end{aligned}

\begin{aligned} = \sin^{- 1} (2 \cos θ \sqrt{1 - \cos^{2} θ}) \\ = \sin^{- 1} (2 \cos θ \sin θ) \\ = \sin^{- 1} (\sin 2 θ) \end{aligned}

u = 2 θ [\begin{array}{l} 2 x = \cos θ \\ θ = \cos^{- 1} (2 x) \end{array}]

\begin{aligned} = 2 \cos^{- 1} (2 x) \\ \frac{d u}{d x} = 2 [\frac{- 1}{\sqrt{1 - (2 x)^{2}}} \times 2] \end{aligned}

= \frac{4}{\sqrt{1 - 4 x^{2}}}

\begin{aligned} v = \sqrt{1 - 4 x^{2}} \\ \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - 4 x^{2}}} (- 8 x) \end{aligned}

= \frac{- 4 x}{\sqrt{1 - 4 x^{2}}}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{4}{\sqrt{1 - 4 x^{2}}}}{\frac{- 4 x}{\sqrt{1 - 4 x^{2}}}} = \frac{- 1}{x}

Differentiation exercise 10.8 question 6

Answer:

\frac{1}{4}

Hint:

Let u = \tan^{- 1} [\frac{\sqrt{1 + x^{2}} - 1}{x}]

v = \sin^{- 1} [\frac{2 x}{1 + x^{2}}]

Given:

\tan^{- 1} [\frac{\sqrt{1 + x^{2}} - 1}{x}] w.r.t \sin^{- 1} [\frac{2 x}{1 + x^{2}}]

- 1 < x < 1, x \neq 0

Explanation:

Let x = \tan θ

u = \tan^{- 1} [\frac{\sqrt{1 + x^{2}} - 1}{x}]

\begin{aligned} = \tan^{- 1} [\frac{\sqrt{1 + \tan^{2} θ} - 1}{\tan θ}] \\ = \tan^{- 1} [\frac{\sqrt{\sec^{2} θ} - 1}{\tan θ}] \\ = \tan^{- 1} [\frac{\sec θ - 1}{\tan θ}] \end{aligned}

\begin{aligned} = \tan^{- 1} [\frac{\frac{1}{\cos θ} - 1}{\frac{\sin θ}{\cos θ}}] \\ = \tan^{- 1} [\frac{1 - \cos θ}{\sin θ}] \end{aligned}

.......................(1)

\begin{aligned} Now \cos 2 θ = 1 - 2 \sin^{2} θ \\ 2 \sin^{2} θ = 1 - \cos 2 θ \\ 2 \sin^{2} \frac{θ}{2} = 1 - \cos θ \end{aligned}

\begin{aligned} \sin 2 θ = 2 \sin θ \cos θ \\ \sin θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2} \end{aligned}

Put in (1)

\begin{aligned} u = \tan^{- 1} [\frac{2 \sin^{2} \frac{θ}{2}}{2 \sin \frac{θ}{2} \cos \frac{θ}{2}}] \\ u = \tan^{- 1} [\tan \frac{θ}{2}] \end{aligned}

Now,

\begin{aligned} - 1 < x < 1 \\ - 1 < \tan θ < 1 \\ - \frac{π}{4} < θ < \frac{π}{4} \end{aligned}

.........(2)

\begin{aligned} - \frac{π}{8} < \frac{θ}{2} < \frac{π}{8} \\ u = \frac{θ}{2} as \frac{θ}{2} \in (- \frac{π}{8}, \frac{π}{8}) \end{aligned}

\begin{aligned} u = \frac{\tan^{- 1} x}{2} \\ \frac{d u}{d x} = \frac{1}{2} [\frac{1}{1 + x^{2}}] \end{aligned}

= \frac{1}{2 (1 + x^{2})}

\begin{aligned} v = \sin^{- 1} [\frac{2 x}{1 + x^{2}}] \\ x = \tan θ \\ v = \sin^{- 1} [\frac{2 (\tan θ)}{1 + \tan^{2} θ}] \end{aligned}

= \sin^{- 1} (\sin 2 θ)

- \frac{π}{4} < θ < \frac{π}{4}

........From (2)

- \frac{π}{2} < 2 θ < \frac{π}{2}

v = \sin^{- 1} (\sin 2 θ) 2 θ \in (- \frac{π}{2}, \frac{π}{2})

= 2 θ

\begin{aligned} v = 2 \tan^{- 1} x \\ \frac{d v}{d x} = \frac{2}{1 + x^{2}} \end{aligned}

\begin{aligned} \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{1}{2 (1 + x^{2})}}{\frac{2}{1 + x^{2}}} \\ = \frac{1}{4} \end{aligned}

Differentiation exercise 10.8 question 7(i)

Answer: 2
Hint:

Let u = \sin^{- 1} [2 x \sqrt{1 - x^{2}}]

v = \sec^{- 1} [\frac{1}{\sqrt{1 - x^{2}}}]

Given:

\sin^{- 1} [2 x \sqrt{1 - x^{2}}] w.r.t \sec^{- 1} [\frac{1}{\sqrt{1 - x^{2}}}]

x \in (0, \frac{1}{\sqrt{2}})

Explanation:

Let x = \sin θ

\begin{aligned} u = \sin^{- 1} [2 x \sqrt{1 - x^{2}}] \\ v = \sec^{- 1} [\frac{1}{\sqrt{1 - x^{2}}}] \end{aligned}

u = \sin^{- 1} [2 \sin θ \sqrt{1 - \sin^{2} θ}]

\begin{aligned} = \sin^{- 1} [2 \sin θ \cos θ] \\ = \sin^{- 1} [\sin 2 θ] \end{aligned}

\begin{aligned} x \in (0, \frac{1}{\sqrt{2}}) \\ \sin θ \in (0, \frac{1}{\sqrt{2}}) \\ θ \in (0, \frac{π}{4}) 2 θ \in (0, \frac{π}{2}) \end{aligned}

\begin{aligned} u = \sin^{- 1} [\sin 2 θ] = 2 θ when 2 θ \in (0, \frac{π}{2}) \\ u = 2 \sin^{- 1} x \end{aligned}

= \frac{2}{\sqrt{1 - x^{2}}}

\begin{aligned} v = \sec^{- 1} [\frac{1}{\sqrt{1 - \sin^{2} θ}}] \\ \sec^{- 1} (\frac{1}{\cos θ}) = \sec^{- 1} (\sec θ) \\ = θ when θ \in (0, \frac{π}{4}) \end{aligned}

\begin{aligned} = \sin^{- 1} x \\ v = \sin^{- 1} x \\ \frac{d v}{d x} = \frac{1}{\sqrt{1 - x^{2}}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{2}{\sqrt{1 - x^{2}}}}{\frac{1}{\sqrt{1 - x^{2}}}} = 2

Differentiation exercise 10.8 question 7(ii)

Answer:

- 2

Hint:

Let u = \sin^{- 1} [2 x \sqrt{1 - x^{2}}]

v = \sec^{- 1} [\frac{1}{\sqrt{1 - x^{2}}}]

Given:

\sin^{- 1} [2 x \sqrt{1 - x^{2}}] w.r.t \sec^{- 1} [\frac{1}{\sqrt{1 - x^{2}}}]

x \in (\frac{1}{\sqrt{2}}, 1)

Explanation:

\begin{aligned} x \in (\frac{1}{\sqrt{2}}, 1) \\ \sin θ \in (\frac{1}{\sqrt{2}}, 1) \\ θ \in (\frac{π}{4}, \frac{π}{2}) 2 θ \in (\frac{π}{2}, π) \end{aligned}

\begin{aligned} π - 2 θ \in (0, \frac{π}{2}) \\ u = \sin^{- 1} [\sin 2 θ] when π - 2 θ \in (0, \frac{π}{2}) \\ u = \sin^{- 1} [\sin (π - 2 θ)] \end{aligned}

\begin{aligned} = π - 2 θ \\ u = π - 2 \sin^{- 1} x \\ \frac{d u}{d x} = \frac{- 2}{\sqrt{1 - x^{2}}} \\ v = \sec^{- 1} [\frac{1}{\sqrt{1 - \sin^{2} θ}}] \end{aligned}

\begin{aligned} \sec^{- 1} (\frac{1}{\cos θ}) = \sec^{- 1} (\sec θ) \\ = θ when θ \in (0, \frac{π}{4}) \end{aligned}

\begin{aligned} = \sin^{- 1} x \\ v = \sin^{- 1} x \\ \frac{d v}{d x} = \frac{1}{\sqrt{1 - x^{2}}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{- 2}{\sqrt{1 - x^{2}}}}{\frac{1}{\sqrt{1 - x^{2}}}} = - 2

Differentiation exercise 10.8 question 8

Answer:

\frac{(\cos x)^{\sin x} {\cos x \cdot \log \cos x - \sin x \tan x}}{(\sin x)^{\cos x} {- \sin x \log \sin x + \cos x \cot x}}

Hint:

Let u = (\cos x)^{\sin x}, v = (\sin x)^{\cos x}

Given:

(\cos x)^{\sin x} w.r.t (\sin x)^{\cos x}

Explanation: Apply log on both sides

\begin{aligned} \log u = \log (\cos x)^{\sin x} \\ \log u = \sin x \log (\cos x) \end{aligned}

Differentiate both side w.r.t.

x

\frac{1}{u} \frac{d u}{d x} = \sin x [\frac{1}{\cos x} (- \sin x)] + \log (\cos x) \cos x

= - \sin x \tan x + \cos x \log (\cos x)

\begin{aligned} \frac{d u}{d x} = (\cos x)^{\sin x} {- \sin x \tan x + \cos x \cdot \log \cos x} \\ v = (\sin x)^{\cos x} \end{aligned}

Apply log on both sides

\begin{aligned} \log v = \log (\sin x)^{\cos x} \\ \log v = \cos x \cdot \log (\sin x) \end{aligned}

Differentiate both side w.r.t.

x

\frac{1}{v} \frac{d v}{d x} = \cos x [- \frac{1}{\sin x} (\cos x)] + \log (\sin x) (- \sin x)

= \cos x \cot x - \sin x \log \sin x

\begin{aligned} \frac{d v}{d x} = (\sin x)^{\cos x} {- \sin x \log \sin x + \cos x \cot x} \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} \end{aligned}

= \frac{(\cos x)^{\sin x} {\cos x \cdot \log \cos x - \sin x \tan x}}{(\sin x)^{\cos x} {- \sin x \log \sin x + \cos x \cot x}}

Differentiation exercise 10.8 question 9

Answer: 1
Hint:

Let u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}), v = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

Given:

\sin^{- 1} (\frac{2 x}{1 + x^{2}}) w.r.t \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

0 < x < 1

Explanation:

Let u = \sin^{- 1} (\frac{2 x}{1 + x^{2}})

\begin{aligned} v = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) \\ Let x = \tan θ \\ u = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}) \end{aligned}

\begin{aligned} = \sin^{- 1} (\sin 2 θ) \\ 0 < x < 1 \\ 0 < \tan θ < 1 \\ 0 < θ < \frac{π}{4} \\ 0 < 2 θ < \frac{π}{2} \end{aligned}

\begin{aligned} u = \sin^{- 1} (\sin 2 θ) = 2 θ 2 θ \in (0, \frac{π}{2}) \\ u = 2 \tan^{- 1} x \end{aligned}

\begin{aligned} \frac{d u}{d x} = \frac{2}{1 + x^{2}} \\ v = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) \end{aligned}

\begin{aligned} v = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \\ = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \\ = \cos^{- 1} (\cos 2 θ) \end{aligned}

\begin{aligned} = 2 θ 2 θ \in (0, \frac{π}{2}) \\ v = 2 \tan^{- 1} x \\ \frac{d v}{d x} = \frac{2}{1 + x^{2}} \end{aligned}

\begin{aligned} \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} \\ = \frac{\frac{2}{1 + x^{2}}}{\frac{2}{1 + x^{2}}} = 1 \end{aligned}

Differentiation exercise 10.8 question 10

Answer:

\frac{1}{a x \sqrt{1 + a^{2} x^{2}}}

Hint:

Let u = \tan^{- 1} (\frac{1 + a x}{1 - a x}), v = \sqrt{1 + a^{2} x^{2}}

Given:

\tan^{- 1} (\frac{1 + a x}{1 - a x}) w.r.t \sqrt{1 + a^{2} x^{2}}

Explanation:

Let u = \tan^{- 1} (\frac{1 + a x}{1 - a x})

\begin{aligned} v = \sqrt{1 + a^{2} x^{2}} \\ Let a x = \tan θ \\ u = \tan^{- 1} (\frac{1 + \tan θ}{1 - \tan θ}) \end{aligned}

= \tan^{- 1} (\frac{\tan \frac{π}{4} + \tan θ}{1 - \tan \frac{π}{4} \tan θ}) [\tan \frac{π}{4} = 1]

\begin{aligned} = \tan^{- 1} \tan (\frac{π}{4} + θ) \\ = \frac{π}{4} + θ \end{aligned}

u = \frac{π}{4} + \tan^{- 1} (a x) [\begin{array}{l} a x = \tan θ \\ θ = \tan^{- 1} a x \end{array}]

\begin{aligned} \frac{d u}{d x} = \frac{1}{1 + (a x)^{2}} [a \times 1] \\ = \frac{a}{1 + a^{2} x^{2}} \\ v = \sqrt{1 + a^{2} x^{2}} \end{aligned}

\begin{aligned} \frac{d v}{d x} = \frac{1}{2 \sqrt{1 + a^{2} x^{2}}} \frac{d}{d x} (a^{2} x^{2}) \\ = \frac{2 a^{2} x}{2 \sqrt{1 + a^{2} x^{2}}} = \frac{a^{2} x}{\sqrt{1 + a^{2} x^{2}}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}

= \frac{\frac{a}{1 + a^{2} x^{2}}}{\frac{a^{2} x}{\sqrt{1 + a^{2} x^{2}}}} = \frac{1}{a x \sqrt{1 + a^{2} x^{2}}}

Differentiation exercise 10.8 question 11

Answer: 2
Hint:

Let u = \sin^{- 1} (2 x \sqrt{1 - x^{2}}), v = \tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}})

Given:

\sin^{- 1} (2 x \sqrt{1 - x^{2}}) w.r.t \tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}})

Explanation:

Let u = \sin^{- 1} (2 x \sqrt{1 - x^{2}})

\begin{aligned} v = \tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) \\ Let x = \sin θ \\ u = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) \end{aligned}

\begin{aligned} = \sin^{- 1} (2 \cos θ \sin θ) \\ = \sin^{- 1} (\sin 2 θ) \end{aligned}

\begin{aligned} v = \tan^{- 1} (\frac{\sin θ}{\sqrt{1 - \sin^{2} θ}}) \\ = \tan^{- 1} (\frac{\sin θ}{\cos θ}) = \tan^{- 1} (\tan θ) \end{aligned}

Now,

\begin{aligned} \frac{- 1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \\ \frac{- 1}{\sqrt{2}} < \sin θ < \frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} \frac{- π}{4} < θ < \frac{π}{4} \\ \frac{- π}{2} < 2 θ < \frac{π}{2} \end{aligned}

\begin{aligned} u = \sin^{- 1} (\sin 2 θ) \\ = \sin^{- 1} (\sin 2 θ) \\ = 2 θ \\ = 2 \sin^{- 1} x \end{aligned}

\begin{aligned} \frac{d u}{d x} = \frac{2}{\sqrt{1 - x^{2}}} \\ v = \tan^{- 1} (\tan θ) \\ = θ \\ v = \sin^{- 1} x \end{aligned}

\begin{aligned} \frac{d v}{d x} = \frac{1}{\sqrt{1 - x^{2}}} \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} \end{aligned}

= \frac{\frac{2}{\sqrt{1 - x^{2}}}}{\frac{1}{\sqrt{1 - x^{2}}}} = 2

Differentiation exercise 10.8 question 12

Answer: 1
Hint:

: Let u = \tan^{- 1} (\frac{2 x}{1 - x^{2}}), v = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

Given:

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) w.r.t \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})

0 < x < 1

Explanation:

Let u = \tan^{- 1} (\frac{2 x}{1 - x^{2}})

\begin{aligned} v = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) \\ Let x = \tan θ \\ u = \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) \\ = \tan^{- 1} (\tan 2 θ) \end{aligned}

\begin{aligned} v = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) \\ v = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \\ = \cos^{- 1} (\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}) \end{aligned}

\begin{aligned} = \cos^{- 1} (\cos 2 θ) \\ Now, \\ 0 < x < 1 \\ 0 < \tan θ < 1 \end{aligned}

\begin{aligned} 0 < θ < \frac{π}{4} \\ 0 < 2 θ < \frac{π}{2} \\ u = \tan^{- 1} (\tan 2 θ) \end{aligned}

\begin{array}{ll} = 2 θ & 2 θ \in (0, \frac{π}{2}) \\ = 2 \tan^{- 1} x & (\begin{matrix} \tan θ = x \\ θ = \tan^{- 1} x \end{matrix}) \end{array}

\begin{aligned} v = \cos^{- 1} (\cos 2 θ) \\ = 2 θ 2 θ \in (0, \frac{π}{2}) \\ = 2 \tan^{- 1} x \end{aligned}

\begin{aligned} \frac{d u}{d x} = \frac{2}{1 + x^{2}} \\ \frac{d v}{d x} = \frac{2}{1 + x^{2}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}}

= \frac{\frac{2}{1 + x^{2}}}{\frac{2}{1 + x^{2}}} = 1

Differentiation exercise 10.8 question 13

Answer: $\frac{\sqrt{1 - x^{2}}}{3 (1 + x^{2})}$

Hint:

Let u = \tan^{- 1} (\frac{x - 1}{x + 1}), v = \sin^{- 1} (3 x - 4 x^{3})

Given:

\tan^{- 1} (\frac{x - 1}{x + 1}) w.r.t \sin^{- 1} (3 x - 4 x^{3})

\frac{- 1}{2} < x < \frac{1}{2}

Explanation:

Let u = \tan^{- 1} (\frac{x - 1}{x + 1})

\begin{aligned} v = \sin^{- 1} (3 x - 4 x^{3}) \\ u = \tan^{- 1} (\frac{x - 1}{x + 1}) \\ = \tan^{- 1} x - \tan^{- 1} 1 \\ = \tan^{- 1} x - \frac{π}{4} \end{aligned}

\begin{aligned} \frac{d u}{d x} = \frac{1}{1 + x^{2}} \\ v = \sin^{- 1} (3 x - 4 x^{3}) \\ Let x = \sin θ \end{aligned}

\begin{aligned} \frac{- 1}{2} < x < \frac{1}{2} \\ \frac{- 1}{2} < \sin θ < \frac{1}{2} \\ - \frac{π}{6} < θ < \frac{π}{6} \end{aligned}

\begin{aligned} v = \sin^{- 1} (3 \sin θ - 4 \sin^{3} θ) \\ = \sin^{- 1} (\sin 3 θ) \\ = 3 θ \\ v = 3 \sin^{- 1} x \end{aligned}

\begin{aligned} \frac{d v}{d x} = \frac{3}{\sqrt{1 - x^{2}}} \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{1}{1 + x^{2}}}{\frac{3}{\sqrt{1 - x^{2}}}} = \frac{\sqrt{1 - x^{2}}}{3 (1 + x^{2})} \end{aligned}

Differentiation exercise 10.8 question 14

Answer: $\frac{- x \sqrt{x^{2} - 1}}{2}$

Hint:

Let u = \tan^{- 1} (\frac{\cos x}{1 + \sin x}), v = \sec^{- 1} x

Given:

\tan^{- 1} (\frac{\cos x}{1 + \sin x}) w.r.t \sec^{- 1} x

Explanation:

Let u = \tan^{- 1} (\frac{\cos x}{1 + \sin x})

\begin{aligned} v = \sec^{- 1} x \\ u = \tan^{- 1} (\frac{\cos^{2} \frac{x}{2} - \sin^{2} \frac{x}{2}}{\cos^{2} \frac{x}{2} + \sin^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2}}) \end{aligned}

[\begin{array}{l} \cos 2 x = \cos^{2} x - \sin^{2} x \\ \sin 2 x = 2 \sin x \cos x \\ 1 = \cos^{2} x + \sin^{2} x \end{array}]

u = \tan^{- 1} (\frac{(\cos \frac{x}{2} - \sin \frac{x}{2}) (\cos \frac{x}{2} + \sin \frac{x}{2})}{{(\cos \frac{x}{2} + \sin \frac{x}{2})}^{2}})

u = \tan^{- 1} (\frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}})

Divide numerator and denominator by

\cos \frac{x}{2}

u = \tan^{- 1} (\frac{1 - \tan \frac{x}{2}}{1 + \tan \frac{x}{2}})

u = \tan^{- 1} (\frac{\tan \frac{π}{4} - \tan \frac{x}{2}}{1 + \tan \frac{π}{4} \tan \frac{x}{2}})

\begin{aligned} u = \tan^{- 1} (\tan (\frac{π}{4} - \frac{x}{2})) \\ u = \frac{π}{4} - \frac{x}{2} \\ \frac{d u}{d x} = \frac{- 1}{2} \end{aligned}

\begin{aligned} v = \sec^{- 1} x \\ \frac{d v}{d x} = \frac{1}{x \sqrt{x^{2} - 1}} \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{- 1}{2}}{\frac{1}{x \sqrt{x^{2} - 1}}} = \frac{- x \sqrt{x^{2} - 1}}{2} \end{aligned}

Differentiation exercise 10.8 question 15

Answer: 1
Hint:

Let u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}), v = \tan^{- 1} (\frac{2 x}{1 - x^{2}})

Given:

\sin^{- 1} (\frac{2 x}{1 + x^{2}}) w.r.t \tan^{- 1} (\frac{2 x}{1 - x^{2}})

- 1 < x < 1

Explanation:

\begin{aligned} Let u = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \\ v = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) \\ Let x = \tan θ \end{aligned}

\begin{aligned} u = \sin^{- 1} (\frac{2 \tan θ}{1 + \tan^{2} θ}), v = \tan^{- 1} (\frac{2 \tan θ}{1 - \tan^{2} θ}) \\ u = \sin^{- 1} (\sin 2 θ), v = \tan^{- 1} (\tan 2 θ) \end{aligned}

\begin{aligned} - 1 < x < 1 \\ - 1 < \tan θ < 1 \end{aligned}

\begin{aligned} - \frac{π}{4} < θ < \frac{π}{4} \\ - \frac{π}{2} < 2 θ < \frac{π}{2} \end{aligned}

\begin{aligned} u = \sin^{- 1} (\sin 2 θ) = 2 θ \\ v = \tan^{- 1} (\tan 2 θ) = 2 θ \\ u = 2 \tan^{- 1} x \end{aligned}

\begin{aligned} \frac{d u}{d x} = \frac{2}{1 + x^{2}} \\ v = 2 \tan^{- 1} x \end{aligned}

\begin{aligned} \frac{d v}{d x} = \frac{2}{1 + x^{2}} \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{2}{1 + x^{2}}}{\frac{2}{1 + x^{2}}} = 1 \end{aligned}

Differentiation exercise 10.8 question 16

Answer: 3
Hint:

Let u = \cos^{- 1} (4 x^{3} - 3 x), v = \tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{x})

Given:

\cos^{- 1} (4 x^{3} - 3 x) w.r.t \tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{x})

\frac{1}{2} < x < 1

Explanation:

u = \cos^{- 1} (4 x^{3} - 3 x)

\begin{aligned} v = \tan^{- 1} (\frac{\sqrt{1 - x^{2}}}{x}) \\ Let x = \cos θ \end{aligned}

\begin{aligned} u = \cos^{- 1} (4 \cos^{3} θ - 3 \cos θ) \\ u = \cos^{- 1} (\cos 3 θ) \end{aligned}

\begin{aligned} v = \tan^{- 1} (\frac{\sqrt{1 - \cos^{2} θ}}{\cos θ}) \\ v = \tan^{- 1} (\frac{\sin θ}{\cos θ}) = \tan^{- 1} (\tan θ) \end{aligned}

\begin{aligned} \frac{1}{2} < x < 1 \\ \frac{1}{2} < \cos θ < 1 \\ 0 < θ < \frac{π}{3} \end{aligned}

\begin{array}{ll} u = \cos^{- 1} (\cos 3 θ) = 3 θ & θ \in (0, \frac{π}{3}) \\ v = \tan^{- 1} (\tan θ) = θ & θ \in (0, \frac{π}{3}) \end{array}

\begin{aligned} u = 3 \cos^{- 1} x \\ \frac{d u}{d x} = \frac{- 3}{\sqrt{1 - x^{2}}} \\ v = \cos^{- 1} x \end{aligned}

\begin{aligned} \frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \\ \frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{- 3}{\sqrt{1 - x^{2}}}}{\frac{- 1}{\sqrt{1 - x^{2}}}} = 3 \end{aligned}

Differentiation exercise 10.8 question 17

Answer:

\frac{1}{2}

Hint:

Let u = \tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}), v = \sin^{- 1} (2 x \sqrt{1 - x^{2}})

Given:

\tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) w.r.t \sin^{- 1} (2 x \sqrt{1 - x^{2}})

- \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}

Explanation:

u = \tan^{- 1} (\frac{x}{\sqrt{1 - x^{2}}})

\begin{aligned} v = \sin^{- 1} (2 x \sqrt{1 - x^{2}}) \\ Let x = \sin θ \end{aligned}

\begin{aligned} u = \tan^{- 1} (\frac{\sin θ}{\sqrt{1 - \sin^{2} θ}}), \\ u = \tan^{- 1} (\frac{\sin θ}{\sqrt{\cos^{2} θ}}) \\ u = \tan^{- 1} (\tan θ) \end{aligned}

\begin{aligned} v = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) \\ v = \sin^{- 1} (2 \sin θ \cos θ) \\ v = \sin^{- 1} (\sin 2 θ) \end{aligned}

Now

\begin{aligned} - \frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} < \sin θ < \frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} - \frac{π}{4} < θ < \frac{π}{4} \\ - \frac{π}{2} < 2 θ < \frac{π}{2} \end{aligned}

\begin{array}{ll} u = \tan^{- 1} (\tan θ) = θ & θ \in (- \frac{π}{4}, \frac{π}{4}) \\ v = \sin^{- 1} (\sin 2 θ) = 2 θ & 2 θ \in (- \frac{π}{2}, \frac{π}{2}) \end{array}

\begin{aligned} u = \sin^{- 1} x \\ \frac{d u}{d x} = \frac{1}{\sqrt{1 - x^{2}}} \\ v = 2 \sin^{- 1} x \end{aligned}

\frac{d v}{d x} = \frac{2}{\sqrt{1 - x^{2}}}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{1}{\sqrt{1 - x^{2}}}}{\frac{2}{\sqrt{1 - x^{2}}}} = \frac{1}{2}

\frac{d u}{d v} = \frac{1}{2}

Differentiation exercise 10.8 question 18

Answer: 1
Hint:

Let x = \cos θ

Given:

\sin^{- 1} \sqrt{1 - x^{2}} w.r.t \cot^{- 1} (\frac{x}{\sqrt{1 - x^{2}}})

Explanation:

Let x = \cos θ

\begin{aligned} u = \sin^{- 1} \sqrt{1 - x^{2}} \\ = \sin^{- 1} \sqrt{1 - \cos^{2} θ} \\ = \sin^{- 1} (\sin θ) \end{aligned}

\begin{aligned} v = \cot^{- 1} (\frac{x}{\sqrt{1 - x^{2}}}) \\ = \cot^{- 1} (\frac{\cos θ}{\sqrt{1 - \cos^{2} θ}}) \end{aligned}

= \cot^{- 1} (\frac{\cos θ}{\sin θ}) = \cot^{- 1} (\cot θ)

Now

\begin{aligned} 0 < x < 1 \\ 0 < \cos θ < 1 \\ 0 < θ < \frac{π}{2} \end{aligned}

\begin{array}{ll} u = \sin^{- 1} (\sin θ) = θ & θ \in (0, \frac{π}{2}) \\ v = \cot^{- 1} (\cot θ) = θ & 2 θ \in (0, \frac{π}{2}) \end{array}

\begin{aligned} u = \cos^{- 1} x [\begin{matrix} x = \cos θ \\ θ = \cos^{- 1} x \end{matrix}] \\ \frac{d u}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \end{aligned}

\begin{aligned} v = \cos^{- 1} x \\ \frac{d v}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{- 1}{\sqrt{1 - x^{2}}}}{\frac{- 1}{\sqrt{1 - x^{2}}}} = 1

\frac{d u}{d v} = 1

Differentiation exercise 10.8 question 19

Answer:

\frac{- 2}{a x}

Hint:

Let a x = \sin θ

Given:

\sin^{- 1} (2 a x \sqrt{1 - a^{2} x^{2}}) w.r.t \sqrt{1 - a^{2} x^{2}}

\frac{- 1}{\sqrt{2}} < a x < \frac{1}{\sqrt{2}}

Explanation:

Let a x = \sin θ

\begin{aligned} u = \sin^{- 1} (2 a x \sqrt{1 - a^{2} x^{2}}) \\ = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) \\ = \sin^{- 1} (2 \sin θ \cos θ) \\ = \sin^{- 1} (\sin 2 θ) \end{aligned}

\begin{aligned} u = \sin^{- 1} (2 a x \sqrt{1 - a^{2} x^{2}}) \\ = \sin^{- 1} (2 \sin θ \sqrt{1 - \sin^{2} θ}) \\ = \sin^{- 1} (2 \sin θ \cos θ) \\ = \sin^{- 1} (\sin 2 θ) \end{aligned}

Now

\begin{aligned} \frac{- 1}{\sqrt{2}} < a x < \frac{1}{\sqrt{2}} \\ \frac{- 1}{\sqrt{2}} < \sin θ < \frac{1}{\sqrt{2}} \end{aligned}

\begin{aligned} - \frac{π}{4} < θ < \frac{π}{4} \\ - \frac{π}{2} < 2 θ < \frac{π}{2} \end{aligned}

\begin{aligned} u = \sin^{- 1} (\sin 2 θ) = 2 θ 2 θ \in (- \frac{π}{2}, \frac{π}{2}) \\ = 2 \sin^{- 1} a x \end{aligned}

\begin{aligned} \frac{d u}{d x} = \frac{2}{\sqrt{1 - a^{2} x^{2}}} (a) \\ = \frac{2 a}{\sqrt{1 - a^{2} x^{2}}} \end{aligned}

\begin{aligned} v = \sqrt{1 - a^{2} x^{2}} \\ \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - a^{2} x^{2}}} (- a^{2} 2 x) \\ \frac{d v}{d x} = \frac{- a^{2} x}{\sqrt{1 - a^{2} x^{2}}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{\frac{2 a}{\sqrt{1 - a^{2} x^{2}}}}{\frac{- a^{2} x}{\sqrt{1 - a^{2} x^{2}}}} = \frac{- 2}{a x}

Differentiation exercise 10.8 question 20

Answer:

\frac{\sqrt{1 - x^{2}}}{x (1 + x^{2})}

Hint:

\tan^{- 1} (\frac{a - b}{a + b}) = \tan^{- 1} a - \tan^{- 1} b

Given:

\tan^{- 1} (\frac{1 - x}{1 + x}) w.r.t \sqrt{1 - x^{2}}

- 1 < x < 1

Explanation:

\tan^{- 1} (\frac{1 - x}{1 + x}) = u

\begin{aligned} u = \tan^{- 1} (1) - \tan^{- 1} x \\ u = \frac{π}{4} - \tan^{- 1} x \\ \frac{d u}{d x} = 0 - \frac{1}{1 + x^{2}} = - \frac{1}{1 + x^{2}} \end{aligned}

\begin{aligned} v = \sqrt{1 - x^{2}} \\ \frac{d v}{d x} = \frac{1}{2 \sqrt{1 - x^{2}}} (- 2 x) \\ \frac{d v}{d x} = \frac{- x}{\sqrt{1 - x^{2}}} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{- \frac{1}{1 + x^{2}}}{\frac{- x}{\sqrt{1 - x^{2}}}} = \frac{\sqrt{1 - x^{2}}}{x (1 + x^{2})}

Differentiation exercise 10.8 question 21

Answer:

- 2 e^{- \cos x} \cos x

Hint:

let u = \sin^{2} x, v = e^{\cos x}

Given:

\sin^{2} x w.r.t e^{\cos x}

- 1 < x < 1

Explanation:

u = \sin^{2} x

\begin{aligned} \frac{d u}{d x} = 2 \sin x \cos x \\ v = e^{\cos x} \\ \frac{d v}{d x} = e^{\cos x} (- \sin x) = - \sin x e^{\cos x} \end{aligned}

\frac{d u}{d v} = \frac{\frac{d u}{d x}}{\frac{d v}{d x}} = \frac{2 \sin x \cos x}{- \sin x e^{\cos x}} = - 2 \cos x e^{- \cos x}

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RD Sharma Class 12 Exercise 10.8 Differentiation Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Solutions Chapter 10 Differentiation - Other Exercise

Differentiation Excercise: 10.8

Differentiation exercise 10.8 question 1

Answer: −xx2−12

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Answer: $\frac{- x \sqrt{x^{2} - 1}}{2}$