If #f(x) =tan^2(x/2) # and #g(x) = sqrt(5x-1 #, what is #f'(g(x)) #? Calculus Basic Differentiation Rules Chain Rule 1 Answer Shwetank Mauria Mar 16, 2016 #f'(g(x))=tan(sqrt(5x-1)/2)xxsec^2(sqrt(5x-1)/2)# Explanation: As #f(x)=tan^2(x/2)#, #f'(x)=(d(tan^2(x/2)))/(d(tanx/2))##xx##(d(tanx/2))/(d(x/2))##xx##(d(x/2))/dx# Hence #f'(x)=2tan(x/2)xxsec^2(x/2)xx1/2# or #f'(x)=tan(x/2)xxsec^2(x/2)# Hence #f'(g(x))=tan(g(x)/2)xxsec^2(g(x)/2)# and as #g(x)=sqrt(5x-1)# #f'(g(x))=tan(sqrt(5x-1)/2)xxsec^2(sqrt(5x-1)/2)# 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 1936 views around the world You can reuse this answer Creative Commons License