How do you take the derivative of $tan (sin x)$ $tan(sin x)$ ?

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How do you take the derivative of $tan (sin x)$ $tan(sin x)$ ?

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Answer 1 · 2015-07-15T09:24:53

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

Explanation:

For this we can use the chain rule: $\frac{d}{d x} tan (sin x) = \frac{d sin x}{d x} \frac{d}{d sin x} tan (sin x) = cos x \frac{d}{d sin x} tan (sin x)$ $d/(dx)tan(sinx)=(dsinx)/(dx)d/(dsinx)tan(sinx)=cosxd/(dsinx)tan(sinx)$

$tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ , so $\frac{d}{d x} tan x = \frac{{cos}^{2} x + {sin}^{2} x}{{cos}^{2} x} = \frac{1}{{cos}^{2} x} = {sec}^{2} x$ $d/(dx)tanx=(cos^2x+sin^2x)/cos^2x=1/cos^2x=sec^2x$ , using the quotient rule and some trigonometric identities.

Therefore we find $\frac{d}{d x} tan (sin x) = cos x {sec}^{2} (sin x)$ $d/(dx)tan(sinx)=cosxsec^2(sinx)$ .

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How do you take the derivative of $tan (sin x)$ $tan(sin x)$ ?

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How do you take the derivative of tan(sinx)tan(sin x)?

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Explanation:

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How do you take the derivative of $tan (sin x)$ $tan(sin x)$ ?