How do you find the derivative of #y = sqrt(2x - x^2)#? Calculus Basic Differentiation Rules Chain Rule 1 Answer GiĆ³ Apr 29, 2015 Write it as: #y=(2x-x^2)^(1/2)# and use the Chain Rule to get: #y'=1/2(2x-x^2)^(1/2-1)*(2-2x)=# #=(1-x)/sqrt(2x-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 10487 views around the world You can reuse this answer Creative Commons License