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

1 Answer
mason m
Jan 28, 2016

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))

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