How do you differentiate #f(x)=tane^(4x)# using the chain rule.?

1 Answer
Dec 23, 2015

Answer:

#f'(x)=4e^(4x)sec^2(e^(4x))#

Explanation:

Treat #f(x)# as #tan(u)#, where #u=e^(4x)#.

According to the chain rule,

#d/dx[tan(u)]=sec^2(u)*u'#

So, #f'(x)=sec^2(e^(4x))*d/dx[e^(4x)]#

To find #d/dx[e^(4x)]#, find #d/dx[e^v]# when #v=4x#.

#d/dx[e^v]=e^v*v'#

#d/dx[e^(4x)]=e^(4x)*d/dx[4x]#

#=>4e^(4x)#

Thus,

#f'(x)=4e^(4x)sec^2(e^(4x))#