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

1 Answer
Dec 20, 2016

Let #y = tanu# and #u = sinx#. To apply the chain rule, we must differentiate both.

I'll start by showing you #u = sinx# using first principles so that you get an understanding of how this works.

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

Use the sum expansion formula #sin(A + B) = sinAcosB + cosAsinB#.

#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 #lim_(x-> 0) sinx/x = 1 and (1 - cosx)/x = 0# to evaluate.

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

#u' = cosx#

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

Meanwhile, #y= tanu# can be written as #y = sinu/cosu#. This function, being of the form #f(x) = (g(x))/(h(x))#, can be differentiated using the quotient rule, where #color(magenta)(f'(x) = (g'(x) xx h(x)- g(x) xx h'(x))/(h(x))^2#.

You will find the derivative of #cosx# to be #-sinx#. Hence,

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

Use the pythagorean identity #cos^2theta+ sin^2theta = 1#:

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

#y' = 1/cos^2u#

Use the identity #1/costheta = sectheta#:

#y' = sec^2u#

Finally, we must determine the derivative of the entire function using the chain rule. This states that #color(magenta)(dy/dx = dy/(du) xx (du)/dx)#.

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

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

Hopefully this helps!