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

3 Answers
Mar 11, 2018

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)

Mar 11, 2018

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)

Mar 11, 2018

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)