Application of Derivatives 1 Question 20
20. Find the equation of the normal to the curve $y=(1+x)^{y}+\sin ^{-1}\left(\sin ^{2} x\right)$ at $x=0$.
(1993, 4M)
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Solution:
- Given, $y=(1+x)^{y}+\sin ^{-1}\left(\sin ^{2} x\right)$
Let $y=u+v$, where $u=(1+x)^{y}, v=\sin ^{-1}\left(\sin ^{2} x\right)$.
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
$\log _e u=y \log _e(1+x)$
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{u} \frac{d u}{d x}=\frac{y}{1+x}+\frac{d y}{d x}{\log _e(1+x) } \\ & \Rightarrow \quad \frac{d u}{d x}=(1+x)^{y} \frac{y}{1+x}+\frac{d y}{d x} \log _e(1+x) \end{aligned} $$
Again, $\quad v=\sin ^{-1}\left(\sin ^{2} x\right)$
$\Rightarrow \quad \sin v=\sin ^{2} x$
$\Rightarrow \quad \cos v \frac{d v}{d x}=2 \sin x \cos x$
$\Rightarrow \quad \frac{d v}{d x}=\frac{1}{\cos v}(2 \sin x \cos x)$
$\Rightarrow \quad \frac{d v}{d x}=\frac{2 \sin x \cos x}{\sqrt{1-\sin ^{2} v}}=\frac{2 \sin x \cos x}{\sqrt{1-\sin ^{4} x}}$
From Eq. (i),
$$ \begin{aligned} \frac{d y}{d x} & =(1+x)^{y} \frac{y}{1+x}+\frac{d y}{d x} \log _e(1+x)+\frac{2 \sin x \cos x}{\sqrt{1-\sin ^{4} x}} \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{y(1+x)^{y-1}+2 \sin x \cos x / \sqrt{1-\sin ^{4} x}}{1-(1+x)^{y} \log _e(1+x)} \end{aligned} $$
At $x=0$,
$$ \begin{aligned} y & =(1+0)^{y}+\sin ^{-1} \sin (0)=1 \\ \therefore \quad \frac{d y}{d x} & =\frac{1(1+0)^{1-1}+2 \sin 0 \cdot \cos 0 / \sqrt{\left(1-\sin ^{4} 0\right)}}{1-(1+0)^{1} \log _e(1+0)} \\ \Rightarrow \quad \frac{d y}{d x} & =1 \end{aligned} $$
Again, the slope of the normal is
$$ m=-\frac{1}{d y / d x}=-1 $$
Hence, the required equation of the normal is
$$ \begin{aligned} y-1 & =(-1)(x-0) \\ \text { i.e. } \quad y+x-1 & =0 \end{aligned} $$
