SATHEE: Application of Derivatives 1 Question 20

Application of Derivatives 1 Question 20

20. Find the equation of the normal to the curve $y = (1 + x)^{y} + \sin^{- 1} (\sin^{2} x)$ at $x = 0$ .

(1993, 4M)

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Solution:

Given, $y = (1 + x)^{y} + \sin^{- 1} (\sin^{2} x)$

Let $y = u + v$ , where $u = (1 + x)^{y}, v = \sin^{- 1} (\sin^{2} x)$ .

On differentiating w.r.t. $x$ , we get

$\frac{d y}{d x} = \frac{d u}{d x} + \frac{d v}{d x}$

Now,

$u = (1 + x)^{y}$

On taking logarithm both sides, we get

$\log_{e} u = y \log_{e} (1 + x)$

$\begin{aligned} \Rightarrow \frac{1}{u} \frac{d u}{d x} = \frac{y}{1 + x} + \frac{d y}{d x} \log_{e} (1 + x) \\ \Rightarrow \frac{d u}{d x} = (1 + x)^{y} \frac{y}{1 + x} + \frac{d y}{d x} \log_{e} (1 + x) \end{aligned}$

Again, $v = \sin^{- 1} (\sin^{2} x)$

$\Rightarrow \sin v = \sin^{2} x$

$\Rightarrow \cos v \frac{d v}{d x} = 2 \sin x \cos x$

$\Rightarrow \frac{d v}{d x} = \frac{1}{\cos v} (2 \sin x \cos x)$

$\Rightarrow \frac{d v}{d x} = \frac{2 \sin x \cos x}{\sqrt{1 - \sin^{2} v}} = \frac{2 \sin x \cos x}{\sqrt{1 - \sin^{4} x}}$

From Eq. (i),

$\begin{aligned} \frac{d y}{d x} & = (1 + x)^{y} \frac{y}{1 + x} + \frac{d y}{d x} \log_{e} (1 + x) + \frac{2 \sin x \cos x}{\sqrt{1 - \sin^{4} x}} \\ \Rightarrow \frac{d y}{d x} & = \frac{y (1 + x)^{y - 1} + 2 \sin x \cos x / \sqrt{1 - \sin^{4} x}}{1 - (1 + x)^{y} \log_{e} (1 + x)} \end{aligned}$

At $x = 0$ ,

$\begin{aligned} y & = (1 + 0)^{y} + \sin^{- 1} \sin (0) = 1 \\ ∴ \frac{d y}{d x} & = \frac{1 (1 + 0)^{1 - 1} + 2 \sin 0 \cdot \cos 0 / \sqrt{(1 - \sin^{4} 0)}}{1 - (1 + 0)^{1} \log_{e} (1 + 0)} \\ \Rightarrow \frac{d y}{d x} & = 1 \end{aligned}$