Question

How do you differentiate `f(x)=sec4x^5`?

Answer 1

THe answer is `=20x^4sec4x^5tan4x^5`

Explanation:

Firtst, we need rhe derivative of `secx`
We use `(u/v)'=(u'v-uv')/v^2`
`secx =1/cosx`
So `(secx)'=(1/cosx)'=(0*cosx-1*(-sinx))/cos^2x`
`=sinx/cos^2x=secxtanx`
`:.` `f'(x)=(sec4x^5)'=sec4x^5tan4x^5*(4x^5)'`
`=sec4x^5tan4x^5*20x^4`