Question

How do you differentiate `f(x)=sec^2(e^(x^4) )` using the chain rule?

Answer 1

`f^'(x)=8x^3(e^(x^(4)))sec^2(e^(x^(4)))tan(e^(x^(4)))`

Explanation:

Chain Rule:
`color(red)(d/(dx)(g@f(x))=g^'(f(x))*f^'(x))`
`f(x)=sec^2(e^(x^(4)))=[color(red)(sec(e^(x^(4))))]^2`
`f^'(x)=2sec(e^(x^(4)))*d/(dx)color(red)((seccolor(blue)((e^(x^(4)))))`
`f^'(x)=2sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))d/(dx)color(blue)((e^(x^4))`
`f^'(x)=2sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))(e^(x^(4)))d/(dx)(color(red)(x^4))`
`f^'(x)=2sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))(e^(x^(4)))(4x^3)`
`f^'(x)=8x^3(e^(x^(4)))sec(e^(x^(4)))sec(e^(x^(4)))tan(e^(x^(4)))`
`f^'(x)=8x^3(e^(x^(4)))sec^2(e^(x^(4)))tan(e^(x^(4)))`