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

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Answer 1 · 2015-12-23T17:39:32

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

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))#