Question

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

Answer 1

`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))`