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

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Answer 1 · 2015-07-15T09:24:53

$d/(dx)tan(sinx)=cosxsec^2(sinx)$

For this we can use the chain rule: $d/(dx)tan(sinx)=(dsinx)/(dx)d/(dsinx)tan(sinx)=cosxd/(dsinx)tan(sinx)$

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

Therefore we find $d/(dx)tan(sinx)=cosxsec^2(sinx)$ .

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