Question

deferentation of y=tan√sin√x​

Answer 1

Answer:

Answer is in the attachment.

Hope it helps!!!!

Answer 2

Solution :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d\left[tan\left(\sqrt {sin(\sqrt{x})}\right)\right]}{d[\sqrt{sin\sqrt{x})}}}$

$\textsf{By using the chain rule of differentiation, we get :} \\ \\ \underline{\sf{\bigstar\:Chain\:rule\:of\: differentiation\: :-}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

$\sf{Here,}$

• $\sf{y = tan\left(\sqrt{sin(\sqrt{x})}\right)}$
• $\sf{u = \sqrt{sin(\sqrt{x}})}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d\left[tan\left(\sqrt {sin(\sqrt{x})}\right)\right]}{d\left[\sqrt{sin(\sqrt{x})}\right]}\cdot\dfrac{d\left[\sqrt{sin(\sqrt{x})}\right]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot\dfrac{d\left[\sqrt{sin(\sqrt{x})}\right]}{dx}}$

$\textsf{By using the chain rule of differentiation, we get :} \\ \\ \underline{\sf{\bigstar\:Chain\:rule\:of\: differentiation\: :-}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

$\sf{Here,}$

1. $\sf{y = \sqrt{sin(\sqrt{x})}}$

• $\sf{u = sin(\sqrt{x})}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot\dfrac{d\left[\sqrt{sin(\sqrt{x})}\right]}{d\left[sin(\sqrt{x})\right]}\cdot\dfrac{d\left[sin(\sqrt{x})\right]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot \dfrac{1}{2}sin(\sqrt{x})^{\left(\tiny{\dfrac{1}{2} - 1}\right)}\cdot\dfrac{d\left[sin(\sqrt{x})\right]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot \dfrac{1}{2}sin(\sqrt{x})^{-\tiny{\dfrac{1}{2}}}\cdot\dfrac{d\left[sin(\sqrt{x})\right]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot \dfrac{1}{2\sqrt{sin(\sqrt{x})}}\cdot\dfrac{d\left[sin(\sqrt{x})\right]}{dx}}$

$\textsf{By using the chain rule of differentiation, we get :} \\ \\ \underline{\sf{\bigstar\:Chain\:rule\:of\: differentiation\: :-}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

$\sf{Here,}$

• $\sf{y = sin(\sqrt{x})}$
• $\sf{u = \sqrt{x}}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot \dfrac{1}{2\sqrt{sin(\sqrt{x})}}\cdot\dfrac{d\left[sin(\sqrt{x})\right]}{d(\sqrt{x})}\cdot \dfrac{d(\sqrt{x})}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot \dfrac{1}{2\sqrt{sin(\sqrt{x})}}\cdot cos(\sqrt{x})\cdot \dfrac{1}{2}x^{\left(\tiny{\dfrac{1}{2} - 1}\right)}}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot \dfrac{1}{2\sqrt{sin(\sqrt{x})}}\cdot cos(\sqrt{x})\cdot \dfrac{1}{2}x^{-\tiny{\dfrac{1}{2}}}}$

$:\implies \sf{\dfrac{dy}{dx} = sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]\cdot \dfrac{1}{2\sqrt{sin(\sqrt{x})}}\cdot cos(\sqrt{x})\cdot \dfrac{1}{2\sqrt{x}}}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]cos(\sqrt{x})}{4\sqrt{sin(\sqrt{x})}\sqrt{x}}}$

$\boxed{\underline{\therefore \sf{\dfrac{dy}{dx} = \dfrac{sec^{2}\left[\sqrt{sin(\sqrt{x}})\right]cos(\sqrt{x})}{4\sqrt{sin(\sqrt{x})}\sqrt{x}}}}}$