Question #42f0e Calculus Basic Differentiation Rules Chain Rule 1 Answer Anjali G Apr 29, 2017 #y'=-sinx lnsqrtx+frac{cosx}{2x}# Explanation: #y=(cosx)(lnsqrtx)# Using the product rule: #y'=(-sinx)(lnsqrtx)+(cosx)(frac{1/2x^(-1/2)}{x^(1/2)})# #y'=-sinx lnsqrtx+frac{cosx}{2x}# 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 932 views around the world You can reuse this answer Creative Commons License