How do you differentiate #y=sin x^2#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer Sonnhard Jun 4, 2018 #y'=cos(x^2)*2x# Explanation: If you mean #y=sin(x^2)# then we get by the chain rule #y'=cos(x^2)2x# Answer link Related questions What is the derivative of #y=cos(x)# ? What is the derivative of #y=tan(x)# ? How do you find the 108th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x)# from first principle? How do you find the derivative of #y=cos(x^2)# ? How do you find the derivative of #y=e^x cos(x)# ? How do you find the derivative of #y=x^cos(x)#? How do you find the second derivative of #y=cos(x^2)# ? How do you find the 50th derivative of #y=cos(x)# ? How do you find the derivative of #y=cos(x^2)# ? See all questions in Derivative Rules for y=cos(x) and y=tan(x) Impact of this question 1323 views around the world You can reuse this answer Creative Commons License