Question

y=sin(tanx) find dy/dx
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Answer 1

Solution :

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

By applying the chain rule of differentiation, we get :

$\underline{\sf{\bigstar\:Chain\: rule\:of\: differentiation :-}} \\ \\ :\implies\sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}}$

Here,

• y = sin(tan(x))
• u = tan(x)

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

$:\implies\sf{\dfrac{dy}{dx} = cos(tan(x))\times sec^{2}(x)}$

$\boxed{\therefore\sf{\dfrac{dy}{dx} = cos(tan(x)) sec^{2}(x)}}$

Hence, the derivative of sin(tan(x)) is cos(tan(x))sec²(x).