Question

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

Answer 1

`f'(x)=(e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4)))/(2sqrtx)`

Explanation:

The chain rule states that `d/dx[sec(u)]=u'*sec(u)tan(u)`, and here `u=e^(sqrtx-4)`.

`f'(x)=d/dx[e^(sqrtx-4)]*sec(e^(sqrtx-4))tan(e^(sqrtx-4))`

To find this derivative, use the rule that `d/dx[e^u]=u'*e^u`, and this time `u=sqrtx-4`.

`f'(x)=d/dx[sqrtx-4]*e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4))`

To find `d/dx[sqrtx-4]`, use the power rule, in that `d/dx[x^(1/2)-4]=1/2x^(-1/2)=1/(2sqrtx)`. This gives

`f'(x)=(e^(sqrtx-4)sec(e^(sqrtx-4))tan(e^(sqrtx-4)))/(2sqrtx)`