How do you differentiate #f(x)=sin(1/(3x-1))# using the chain rule?

1 Answer
May 14, 2016

Answer:

#(-3cos(1/(3x-1)))/(3x-1)^2#

Explanation:

differentiate using the#color(blue)" chain rule"#

#d/dx[f(g(x))]=f'(g(x)).g'(x)color(red)" (A)"#
#"-------------------------------------------"#

#f(g(x))=sin(1/(3x-1))rArrf'(g(x))=cos(1/(3x-1))#

and#g(x)=1/(3x-1)=(3x-1)^-1 #

#rArrg'(x)=-1(3x-1)^-2 .3=-3(3x-1)^-2=(-3)/(3x-1)^2#
#"-----------------------------------------------------"#
Substitute these values into#color(red)" (A)"#

#f'(x)=cos(1/(3x-1))xx(-3)/(3x-1)^2=(-3cos(1/(3x-1)))/(3x-1)^2#