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

3 Answers
Mar 11, 2018

Answer:

#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

Answer:

#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

Answer:

#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)#