How do you find the derivative of the function #y = sin(tan(5x))#?

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Answer 1 · 2016-05-24T07:59:09

#\cos (\tan (5x))\sec ^2(5x)5#

#\frac{d}{dx}(\sin (\tan (5x)))#

Applying chain rule,
#\frac{df(u)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}#

Let #tan(5x)# = u

We know,
#\frac{d}{du}(\sin(u))=\cos(u)#

#\frac{d}{dx}(\tan(5x))=\sec ^2(5x)5#

So,
#=\cos(u)\sec ^2(5x)5#

Substituting back,#tan(5x)# = u

#=\cos(\tan(5x))\sec ^2(5x)5#