Question

If $x = \sqrt{a^{sin^{-1}t}}$,$y =\sqrt{a^{cos^{-1}t}}$, show that $\frac{dy}{dx}= -\frac{y}{x}$.


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Answer 1

Step-by-step explanation:

$\sf x = \sqrt{a^{sin^{-1} \ t}}\\\\\\Derivative \ rule:\boxed{\dfrac{d(\sqrt{x})}{dx}=\dfrac{1}{2}*x^{\frac{-1}{2}}=\dfrac{1}{2\sqrt{x}}}$

$\sf \dfrac{d(\sqrt{a^{sin^{-1} \ t}}}{dt}=\dfrac{1}{2\sqrt{a^{sin^{-1} \ t}}}*\dfrac{d(a^{sin^{-1} \ t})}{dt}\\\\\\Derivative \ rule: \boxed{\dfrac{d(a^{x})}{dx}=log \ a *a^{x}}$

   

                      $\sf = \dfrac{1}{2\sqrt{a^{sin^{-1}} \ t}}*a^{sin^{-1} \ t}* log \ a *\dfrac{d(Sin^{-1} \ t)}{dt}$

$Derivative \ rule:\boxed{\dfrac{d(sin^{-1} \ x}{dx}=\dfrac{1}{\sqrt{1-x^2}}}$

                       $\sf = \dfrac{1}{2\sqrt{a^{sin^{-1} \ t}}}*a^{Sin^{-1} \ t}*log \ a*\dfrac{1}{\sqrt{1-x^2}}}}}\\\\ = \dfrac{a^{Sin^{-1} \ t}*log \ a}{2\sqrt{a^{sin^{-1} \ t}}*\sqrt{1-x^2}}$

$\boxed{ \dfrac{a^{sin^{-1} \ t}}{\sqrt{a^{sin^{-1} \ t}}}=\dfrac{\sqrt{a^{sin^{-1} \ t}}*\sqrt{a^{sin^{-1} \ t}}}{\sqrt{a^{sin^{-1} \ t}}} = \sqrt{a^{sin^{-1} \ t}}}$

                         $\sf = \dfrac{a^{sin^{-1} \ t}*log \ a}{2\sqrt{1-x^2}}$

$\sf \dfrac{dy}{dt}=\dfrac{d(a^{cos^{-1} \ t})}{dt}$

     $= \dfrac{1}{2\sqrt{a^{cos^{-1} \ t}}}*a^{cos^{-1} \ t}*log \ a *\dfrac{-1}{\sqrt{1-x^2}}}\\\\\\=\dfrac{(-1)*a^{cos^{-1} \ t}*log \ a}{2*\sqrt{a^{cos^{-1} \ t}}*\sqrt{1-x^2}}$

    $\sf = \dfrac{(-1)*\sqrt{a^{Cos^{-1} \ t}}* log \ a }{2\sqrt{1-x^2}}$

                       

                     $\sf \bf \dfrac{dy}{dx}=\dfrac{dy}{dt} \div \dfrac{dx}{dt}$

                            $\sf \bf = \dfrac{(-1)*\sqrt{a^{cos^{-1} \ t}}*log \ a}{2*\sqrt{1-x^2}} \ \div \dfrac{\sqrt{a^{sin^{-1} \ t}} *log \ a}{2*\sqrt{1-x^2}}\\\\\\=\dfrac{(-1)*\sqrt{a^{cos^{-1} \ t}}*log \ a}{2*\sqrt{1-x^2}} \ * \dfrac{2*\sqrt{1-x^2}}{\sqrt{a^{sin^{-1} \ t}} *log \ a}\\\\= \dfrac{(-1)* \sqrt{a^{cos^{-1} \ t}} }{\sqrt{a^{sin^{-1} \ t}}}\\\\= \dfrac{-y}{x}$