How do you use the chain rule to differentiate #y=root5(x^2-3)/(-x-5)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer Manikandan S. Nov 21, 2017 Separate into components Explanation: #y = (root(5)(x^2-3))/(-x-5)# Let #u = root(5)(x^2-3)# and #v =-1/(x+5)# #y' = udv+vdu# #y' = root(5)(x^2-3)\times1/(x+5)^2-1/(x+5)(1/5)(x^2-3)^(-4/5) 2x# #y' = root(5)(x^2-3)\times1/(x+5)^2(1-(2x(x+5))/(x^2-3))# #y' = root(5)(x^2-3)\times1/(x+5)^2(-(x^2+10x+3))/(x^2-3)# #y' = (root(5)(x^2-3))^4 \times(1+22/(x+5)^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 1330 views around the world You can reuse this answer Creative Commons License