How do you differentiate # y =sqrt(sec ^2x^3# using the chain rule? Calculus Basic Differentiation Rules 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# Answer link Related questions What is the Chain Rule for derivatives? How do you find the derivative of #y= 6cos(x^2)# ? How do you find the derivative of #y=6 cos(x^3+3)# ? How do you find the derivative of #y=e^(x^2)# ? How do you find the derivative of #y=ln(sin(x))# ? How do you find the derivative of #y=ln(e^x+3)# ? How do you find the derivative of #y=tan(5x)# ? How do you find the derivative of #y= (4x-x^2)^10# ? How do you find the derivative of #y= (x^2+3x+5)^(1/4)# ? How do you find the derivative of #y= ((1+x)/(1-x))^3# ? See all questions in Chain Rule Impact of this question 1957 views around the world You can reuse this answer Creative Commons License