Question

Solve question no. 11


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Answer 1

Step-by-step explanation:

refer above one!

hope it helps..

Answer 2

Given Question

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

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

Given that,

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

and

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

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

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

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

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

We know,

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

So, using this, we get

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

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

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

We know,

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

So, using this, we get

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

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

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

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

Hence, Proved

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MORE TO KNOW

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