How do you differentiate #e^((2-x)^2) # using the chain rule? Calculus Basic Differentiation Rules Chain Rule 1 Answer Bdub Nov 1, 2016 #y'=-2(2-x)e^((2-x)^2)# Explanation: #y=e^((2-x)^2)# Use the rule #e^u =e^u *u'# #y'=e^((2-x)^2) 2(2-x)*-1# #y'=-2(2-x)e^((2-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 1401 views around the world You can reuse this answer Creative Commons License