What is the derivative of $tan (sin x)$ $tan(sin x)$ ?

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What is the derivative of $tan (sin x)$ $tan(sin x)$ ?

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Answer 1 · 2016-01-02T16:11:51

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

Explanation:

Use the chain rule.

According to the chain rule,

$\frac{d}{d x} (tan (u)) = {sec}^{u} \cdot \frac{d u}{d x}$ $d/dx(tan(u))=sec^(u)*(du)/dx$

Therefore,

$\frac{d}{d x} (tan (sin x)) = {sec}^{2} (sin x) \cdot \frac{d}{d x} (sin x)$ $d/dx(tan(sinx))=sec^2(sinx)*d/dx(sinx)$

$\Rightarrow {sec}^{2} (sin x) cos x$ $=>sec^2(sinx)cosx$

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What is the derivative of $tan (sin x)$ $tan(sin x)$ ?

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