How do you differentiate $tan (sin x)$ $tan(sin x)$ ?

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Answer 1 · 2018-04-28T14:53:15

${sec}^{2} (sin x) cos x$ $sec^2(sinx)cosx$

Given: $\frac{d}{d x} (tan (sin x))$ $d/dx(tan(sinx))$ .

Here, let $y = tan (sin x)$ $y=tan(sinx)$ .

Use the chain rule, which states that,

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du)*(du)/dx$

Let $u=sinx,:.(du)/dx=cosx$ .

Then $y=tanu,dy/(du)=sec^2u$ .

So, combining our results, we get:

$dy/dx=sec^2u*cosx$

Substituting back $u=sinx$ , we get:

$dy/dx=sec^2(sinx)*cosx$

$=sec^2(sinx)cosx$

How do you differentiate tan(sin x)tan(sinx) tan(sin x)?