How do you differentiate # y =sqrt(sec ^2x^3# using the chain rule?

1 Answer
Gerardina C.
Mar 5, 2017

#=(3x^2sinx^3sqrt(cos^2x^3))/cos^3x^3#

Explanation:

Let's apply #secx=1/cosx# and then differentiate:

#y=sqrt(1/cos^2x^3#

Since

1) #f(x)=sqrt(g(x))->f'(x)=1/(2sqrt(g(x)))*g'(x)#
2)#f(x)=1/(g(x))^2=(g(x))^(-2)->f'(x)=-2(g(x))^-3*g'(x)=(-2)/(g(x))^3*g'(x)#
3)#f(x)=cos(g(x))->f'(x)=-sin(g(x))*g'(x)#
4)#f(x)=x^3->f'(x)=3x^2#

then

#y'=1/(cancel2sqrt(1/cos^2x^3))* (-cancel2/cos^3x^3) *(-sinx^3)*3x^2#

#=(3x^2sinx^3sqrt(cos^2x^3))/cos^3x^3#

Related questions
See all questions in Chain Rule
Impact of this question
2012 views around the world
You can reuse this answer
Creative Commons License