How do you differentiate #tan^2(3-x^2)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Massimiliano Apr 14, 2015 In this way: #y'=2tan(3-x^2) * 1/cos^2(3-x^2) * (-2x)=# #=-4x * tan(3-x^2) * 1/cos^2(3-x^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 1781 views around the world You can reuse this answer Creative Commons License