Question

If `f(x) =-e^(2x-1)` and `g(x) = -3sec^2x^2`, what is `f'(g(x))`?

Answer 1

`f'(g(x))=1/(6x)e^(2x-1)cotx^2cos^2x^2`

Explanation:

As `f(x)=-e^(2x-1)` and `g(x)=-3sec^2x^2`

`(df)/(dx)=-e^(2x-1)xx2=-2e^(2x-1)`

and`(dg)/(dx)=-3xx2secx^2xxsecx^2tanx^2xx2x`

= `-12xtanx^2sec^2x^2`

Now `f'(g(x))=(df)/(dg)=((df)/(dx))/((dg)/(dx))`

= `(-2e^(2x-1))/(-12xtanx^2sec^2x^2)`

= `e^(2x-1)/(6xtanx^2sec^2x^2)`

= `1/(6x)e^(2x-1)cotx^2cos^2x^2`