How do you differentiate #f(x)=sec^4(e^(x^3) ) # using the chain rule?

1 Answer
Jan 28, 2016

Answer:

#f'(x)=12x^2e^(x^3)sec^4(e^(x^3))tan(e^(x^3))#

Explanation:

The first issue is the fourth power. We can deal with it through the application of the chain rule #d/dx(u^4)=4u^3*u'#, where #u=sec(e^(x^3))#, yielding

#f'(x)=4sec^3(e^(x^3))*d/dx(sec(e^(x^3)))#

To differentiate the secant function, use the rule: #d/dx(sec(u))=sec(u)tan(u)*u'#, where now #u=e^(x^3)#. This gives

#f'(x)=4sec^3(e^(x^3))sec(e^(x^3))tan(e^(x^3))*d/dx(e^(x^3))#

Which simplifies to be

#f'(x)=4sec^4(e^(x^3))tan(e^(x^3))*d/dx(e^(x^3))#

All that remains is the differentiation of #e^(x^3)#, to which the chain rule will again be applied: #d/dx(e^u)=e^u*u'#. Now, #u=x^3#, so

#f'(x)=4sec^4(e^(x^3))tan(e^(x^3))*e^(x^3)d/dx(x^3)#

Since #d/dx(x^3)=3x^2#,

#f'(x)=12x^2e^(x^3)sec^4(e^(x^3))tan(e^(x^3))#