Question

How do you differentiate `f(x)=sec(3x^3-x^2 )` using the chain rule?

Answer 1

`f'(x)=(9x^2-2x)sec(3x^3-x^2)tan(3x^3-x^2)`

Explanation:

`f(x)=y=sec(3x^3-x^2)`

chain rule

`f'(x)=(dy)/(dx)=(dy)/(du)(du)/(dx)`

`u=3x^3-x^2=>(du)/(dx)=9x^2-2x`

`y=secu=>(dy)/(du)=secutanu`

`f'(x)=secutanuxx(9x^2-2x)`

`=(9x^2-2x)sec(3x^3-x^2)tan(3x^3-x^2)`

Answer 2

`tan(3x^3-x^2).sec(3x^3-x^2).x(9x-2)`

Explanation:

`y= f(g(x))= sec(3x^3-x^2)`

`(df(g(x)))/dx= (df(g(x)))/(dg(x)) . (dg(x))/dx`

`(d sec(3x^3-x^2))/dx= (d sec(3x^3-x^2))/(d(3x^3-x^2)). (d(3x^3-x^2))/dx`

Because derivative of `sec(x)` is `tan x. sec x` :

`(d sec(3x^3-x^2))/dx = tan(3x^3-x^2).sec(3x^3-x^2) . 9x^2-2x`
`= tan(3x^3-x^2).sec(3x^3-x^2).x(9x-2)`

Answer 3

`f'(x)=(9x^2-2x)sec(3x^3-x^2)tan(3x^3-x^2)`

Explanation:

`•color(white)(x)d/dx(secx)=secxtanx`

`"Differentiate using the "color(blue)"chain rule"`

`"Given "f(x)=g(h(x))" then"`

`f'(x)=g'(h(x))xxh'(x)larrcolor(blue)"chain rule"`

`f(x)=sec(3x^3-x^2)`

`rArrf'(x)=sec(3x^3-x^2)tan(3x^3-x^2)xxd/dx(3x^3-x^2)`

`color(white)(rArrf'(x))=(9x^2-2x)sec(3x^3-x^2)tan(3x^3-x^2)`