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  • Explain Solution R.D.Sharma Class 12 Chapter 10 Differentiation Exercise 10.3 Question 34 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 10  Differentiation  Exercise 10.3 Question 34 Maths Textbook Solution.

Answers (1)

Answer:\frac{dy}{dx}=\frac{2^{x+1}log2}{1+4^{x}}

Hint:

    \frac{d}{dx}\left ( constant \right )=0

\frac{\partial }{\partial x}\left ( x^{n} \right )=nx^{n-1}

Given:

\sin ^{-1}\left ( \frac{2^{x+1}}{1+4^{x}} \right )

Solution:

Let

   y=\sin ^{-1}\left ( \frac{2^{x+1}}{1+4^{x}} \right )

To find the domain we need to find all x such that

-1 \leq \frac{2^{x+1}}{1+4 x} \leq 1

Since the quantity in the middle is always positive, we need to find

all such that \frac{2^{x+1}}{1+4 x} \leq 1

ie, all x such that 2^{x+1}=1+4^{x}

2\leq \frac{1}{2^{x}}+2^{x}, which is true for a x

Hence the function is defined at all real numbers
Putting 2^{x}=\tan \theta

\begin{aligned} &y=\sin ^{-1}\left(\frac{2^{x+1}}{1+4 x}\right) \\ &y=\sin ^{-1}\left(\frac{2^{x} \cdot 2}{1+\left(2^{x}\right)^{2}}\right) \\ &y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right) \\ &\text { Using } \sec ^{2} \theta=1+\tan ^{2} \theta \end{aligned}

\begin{aligned} &\mathrm{y}=\sin ^{-1}\left(\frac{2 \tan \theta}{\sec ^{2} \theta}\right) \\ &\mathrm{y}=\sin ^{-1}\left(2 \frac{\sin \theta}{\cos \theta} \times \cos ^{2} \theta\right) \quad \text \; \; \; \; \; \; \; \; \; \; \; { Since, } \tan \theta=\frac{\sin \theta}{\cos \theta^{\prime} \sec \theta}, \frac{1}{\sec }=\cos \theta \\ &\mathrm{y}=\sin ^{-1}(2 \sin \theta \cos \theta) \\ &\mathrm{U} \sin \mathrm{sin}=\sin 2 \theta=2 \sin \theta \cos \theta \end{aligned}

y=\sin ^{-1}\left ( \sin \theta \right )                                                                                                \left\{\begin{array}{l} \sin ^{-1}(\sin \theta)=\theta \\ \text { if } \theta \in\left[\frac{-1}{2}, \frac{0}{2}\right] \end{array}\right\}

y=2\theta

y=2\tan ^{-1}\left ( 2^{x} \right )                                                                                                      \left\{\begin{array}{l} \text { since } 2^{\mathrm{x}} \quad=\tan \theta \\ \boldsymbol{\theta}=\tan ^{-1}\left(2^{\mathrm{x}}\right) \end{array}\right\}

\begin{aligned} &\frac{d y}{d x} \Rightarrow \frac{d}{d x}\left(2 \tan ^{-1}\left(2^{x}\right)\right) \\ &\frac{d y}{d x}=2 \times \frac{1}{1+\left(2^{x}\right)^{2}} \frac{\partial}{d x}\left(2^{x}\right) \\ &\frac{d y}{d x}=2 \times \frac{1}{1+4^{x}} \cdot\left(2^{x}\right) \log 2 \\ &\frac{d y}{d x} \Rightarrow \frac{2^{x+1} \log 2}{1+4^{x}} \end{aligned}                                                        \text { Since } \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}

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