Question

If

$y = \sqrt{ {h}^{ {cos}^{ - 1t}} }$

and

$x = \sqrt{ {h}^{ {sin}^{ - 1} t} }$

Prove that dy/dx= -y/x

​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:y = \sqrt{ {h}^{ {cos}^{ - 1}t } } - - - - (1)$

and

$\rm :\longmapsto\:x = \sqrt{ {h}^{ {sin}^{ - 1}t } } - - - (2)$

On multiply equation (1) and (2), we get

$\rm :\longmapsto\:xy = \sqrt{ {h}^{ {cos}^{ - 1}t } } \times \sqrt{ {h}^{ {sin}^{ - 1}t } }$

We know,

$\boxed{ \tt{ \: {a}^{m} \times {a}^{n} = {a}^{m + n} \: }}$

So, using this Law of radicals, we get

$\rm :\longmapsto\:xy = \sqrt{ {h}^{ {cos}^{ - 1}t + {sin}^{ - 1} t} }$

We know,

$\boxed{ \tt{ \: {cos}^{ - 1}x + {sin}^{ - 1}x = \frac{\pi}{2} \: }}$

So, using this identity, we get

$\rm :\longmapsto\:xy \: = \: \sqrt{ {\bigg[h\bigg]}^{\dfrac{\pi}{2} } }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}xy \: = \dfrac{d}{dx} \: \sqrt{ {\bigg[h\bigg]}^{\dfrac{\pi}{2} } }$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}uv \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$

and

$\boxed{ \tt{ \: \dfrac{d}{dx}k \: = \: 0 \: }}$

So, using this, we get

$\rm :\longmapsto\:x\dfrac{d}{dx}y \: + \: y\dfrac{d}{dx}x \: = \: 0$

$\rm :\longmapsto\:x\dfrac{dy}{dx} + y \times 1 = 0$

$\rm :\longmapsto\:x\dfrac{dy}{dx} + y = 0$

$\rm :\longmapsto\:x\dfrac{dy}{dx} = - y$

$\bf\implies \:\dfrac{dy}{dx} \: = \: - \: \dfrac{y}{x}$

Hence, Proved

Additional Information :-

$\green{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:y = \sqrt{ {h}^{ {cos}^{ - 1}t } } - - - - (1)$

and

$\rm :\longmapsto\:x = \sqrt{ {h}^{ {sin}^{ - 1}t } } - - - (2)$

On multiply equation (1) and (2), we get

$\rm :\longmapsto\:xy = \sqrt{ {h}^{ {cos}^{ - 1}t } } \times \sqrt{ {h}^{ {sin}^{ - 1}t } }$

We know,

$\boxed{ \tt{ \: {a}^{m} \times {a}^{n} = {a}^{m + n} \: }}$

So, using this Law of radicals, we get

$\rm :\longmapsto\:xy = \sqrt{ {h}^{ {cos}^{ - 1}t + {sin}^{ - 1} t} }$

We know,

$\boxed{ \tt{ \: {cos}^{ - 1}x + {sin}^{ - 1}x = \frac{\pi}{2} \: }}$

So, using this identity, we get

$\rm :\longmapsto\:xy \: = \: \sqrt{ {\bigg[h\bigg]}^{\dfrac{\pi}{2} } }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}xy \: = \dfrac{d}{dx} \: \sqrt{ {\bigg[h\bigg]}^{\dfrac{\pi}{2} } }$

We know,

$\boxed{ \tt{ \: \dfrac{d}{dx}uv \: = \: u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u \: }}$

and

$\boxed{ \tt{ \: \dfrac{d}{dx}k \: = \: 0 \: }}$

So, using this, we get

$\rm :\longmapsto\:x\dfrac{d}{dx}y \: + \: y\dfrac{d}{dx}x \: = \: 0$

$\rm :\longmapsto\:x\dfrac{dy}{dx} + y \times 1 = 0$

$\rm :\longmapsto\:x\dfrac{dy}{dx} + y = 0$

$\rm :\longmapsto\:x\dfrac{dy}{dx} = - y$

$\bf\implies \:\dfrac{dy}{dx} \: = \: - \: \dfrac{y}{x}$

Hence, Proved

Additional Information :-

$\green{\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}}$