Solve this
Question:

If $\sin ^{2} y+\cos x y=k$, find $\frac{d y}{d x}$ at $x=1, y=\frac{\pi}{4}$

Solution:

We are given with an equation $\sin ^{2} y+\cos (x y)=k$, we have to find $\frac{d y}{d x}$ at $x=1, y=\frac{\pi}{4}$ by using the given

equation, so by differentiating the equation on both sides with respect to $x$, we get,

$2 \sin y \cos y \frac{d y}{d x}-\sin (x y)\left[(1) y+x \frac{d y}{d x}\right]=0$

$\frac{d y}{d x}[2 \sin y \cos y-x \sin (x y)]=y \sin (x y)$

$\frac{d y}{d x}=\frac{y \sin (x y)}{2 \sin y \cos y-x \sin (x y)}$

By putting the value of point in the derivative, which is $x=1, y=\frac{\pi}{4}$,

$\frac{\mathrm{dy}}{\mathrm{dx}}(\mathrm{x}=1, \mathrm{y}=\mathrm{r} / 4)=\frac{\frac{\pi}{4} \sin \left(\frac{\pi}{4}\right)}{2 \sin \frac{\pi}{4} \cos \frac{\pi}{4}-(1) \sin \frac{\pi}{4}}$

$\frac{d y}{d x}(x=1, y=\pi / 4)=\frac{\frac{\pi}{4 \sqrt{2}}}{1-\frac{1}{\sqrt{2}}}=\frac{\frac{\pi}{4 \sqrt{2}}}{\frac{\sqrt{2}-1}{\sqrt{2}}}=\frac{\pi}{4(\sqrt{2}-1)}$

Administrator
Solve this
Question:

If $\mathrm{y}=\mathrm{x} \sin ^{-1} \mathrm{x}+\sqrt{1-\mathrm{x}^{2}}$, prove that $\frac{\mathrm{dy}}{\mathrm{dx}}=\sin ^{-1} \mathrm{x}$

Solution:

Given $y=x \sin ^{-1} x+\sqrt{1-x^{2}}$

On differentiating $y$ with respect to $x$, we get

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

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x \sin ^{-1} x\right)+\frac{d}{d x}\left(\sqrt{1-x^{2}}\right)$

$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(x \times \sin ^{-1} x\right)+\frac{d}{d x}\left[\left(1-x^{2}\right)^{\frac{1}{2}}\right]$

We have (uv) $^{\prime}=v u^{\prime}+u v^{\prime}$ (product rule)

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin ^{-1} \mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})+\mathrm{x} \frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)+\frac{\mathrm{d}}{\mathrm{dx}}\left[\left(1-\mathrm{x}^{2}\right)^{\frac{1}{2}}\right]$

We know $\frac{\mathrm{d}}{\mathrm{dx}}\left(\sin ^{-1} \mathrm{x}\right)=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}$ and $\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

$\Rightarrow \frac{d y}{d x}=\sin ^{-1} x \times 1+x \times \frac{1}{\sqrt{1-x^{2}}}+\frac{1}{2}\left(1-x^{2}\right)^{\frac{1}{2}-1} \frac{d}{d x}\left(1-x^{2}\right)$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin ^{-1} \mathrm{x}+\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}+\frac{1}{2}\left(1-\mathrm{x}^{2}\right)^{-\frac{1}{2}}\left[\frac{\mathrm{d}}{\mathrm{dx}}(1)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin ^{-1} \mathrm{x}+\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}+\frac{1}{2 \sqrt{1-\mathrm{x}^{2}}}\left[\frac{\mathrm{d}}{\mathrm{dx}}(1)-\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}\right)\right]$

However, $\frac{d}{d x}\left(x^{2}\right)=2 x$ and derivative of a constant is 0 .

$\Rightarrow \frac{d y}{d x}=\sin ^{-1} x+\frac{x}{\sqrt{1-x^{2}}}+\frac{1}{2 \sqrt{1-x^{2}}}[0-2 x]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin ^{-1} \mathrm{x}+\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}-\frac{2 \mathrm{x}}{2 \sqrt{1-\mathrm{x}^{2}}}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\sin ^{-1} \mathrm{x}+\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}-\frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}$

$\therefore \frac{d y}{d x}=\sin ^{-1} x$

Thus, $\frac{d y}{d x}=\sin ^{-1} x$

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