# How do you find the derivative of #f(x) = tan(sinx)#?

##### 1 Answer

Let

I'll start by showing you

#u' = lim_(h->0) (f(x + h) - f(x))/h#

Use the sum expansion formula

#u' = lim_(h->0) (sin(x + h) - sinx)/h#

#u' = lim_(h->0) (cosxsin h - sinx + sinxcos h)/h#

#u' = lim_(h->0) (cosxsin h - sinx(1 - cos h))/h#

#u' = " "cosx * lim_(h->0) sin h/h - sinx * lim_(h->0) (1 - cos h)/h#

We use the two well known limits

#u' = cosx * 1 - sinx * 0#

#u' = cosx#

A similar process can be used to obtain the derivative of cosine.

Meanwhile,

You will find the derivative of

#y' = (cosu xx cosu - (sinu xx -sinu))/(cosu)^2#

Use the pythagorean identity

#y' = (cos^2u + sin^2u)/cos^2u#

#y' = 1/cos^2u#

Use the identity

#y' = sec^2u#

Finally, we must determine the derivative of the entire function using the chain rule. This states that

#dy/dx = sec^2u xx cosx#

#dy/dx = cosxsec^2(sinx)#

Hopefully this helps!