How do you find the derivative of #y=sin^-1(2x+1)#? Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer James Apr 18, 2018 The answer is #y'=(2)/sqrt(1-(2x+1)^2# Explanation: #d/dx[y]=[1/sqrt(1-(2x+1)^2]*d/dx[2x+1]] # #y'=(2)/sqrt(1-(2x+1)^2# 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 1613 views around the world You can reuse this answer Creative Commons License