Question

If

$x = \sqrt{ {a}^{ {cos}^{ - 1} t} } \: and \: y = \sqrt{ {a}^{ {sin}^{ - 1} t} } \: prove \: that \: \frac{dy}{dx} = - \frac{y}{x}$

explain all steps ​

Answer 1

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

Given parametric functions are

$\rm \: x = \sqrt{ {a}^{ {cos}^{ - 1}t } } \: - - - (1)$

and

$\rm \: y = \sqrt{ {a}^{ {sin}^{ - 1}t } } \: - - - (2)$

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

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

We know,

$\boxed{\sf{ \: {x}^{m} \times {y}^{m} \: = \: {(xy)}^{m} \: }}$

So, using this identity, we get

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

We know,

$\boxed{\sf{ \: {x}^{m} \times {x}^{n} \: = \: {x}^{m + n} \: }}$

So, using this identity, we get

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

We know,

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

So, using this identity, we get

$\rm \: xy \: = \: \sqrt{ {\bigg(a\bigg) }^{\dfrac{\pi}{2}} }$

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

$\rm \: \dfrac{d}{dx}(xy) \: = \: \dfrac{d}{dx}\sqrt{ {\bigg(a\bigg) }^{\dfrac{\pi}{2}} }$

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

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

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

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

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

Hence, Proved

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Additional Information

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

Answer 2

Refer the given attachment