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

1 Answer
Oct 19, 2017

Answer:

#f'(x)=-(3(3x)^(-3/2))/2sec((3x)^(-1/2))tan((3x)^(-1/2))=-(3sec(1/sqrt(3x))tan(1/sqrt(3x)))/(2sqrt(3x)^3)#

Explanation:

The equation given can be simplified as #f(x)=sec((3x)^(-1/2))#

If #f(x)=sec(g(x))#, then #f'(x)=g'(x)sec(g(x))tan(g(x))#

If #g(x)=(h(x))^n#, then #g'(x)=h(x)*n(h(x))^(n-1)#

The derivative of #3x# is#3#.

#-1/2-1=-3/2#

#g'(x)=3*-1/2(3x)^(-3/2)=-(3(3x)^(-3/2))/2#

So, #f'(x)=-(3(3x)^(-3/2))/2sec((3x)^(-1/2))tan((3x)^(-1/2))=-(3sec(1/sqrt(3x))tan(1/sqrt(3x)))/(2sqrt(3x)^3)#