What is the derivative of #e^((lnx)^2)#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Base e 1 Answer Tom Apr 4, 2015 #u = (ln(x))^2# Derivate will be #u'*e^u# So : # ((ln(x))^2)' = (u^n)' = n*u'u^(n-1)# here #n = 2# #((ln(x))^2)' = 2*1/x*ln(x) = (2ln(x))/x# Finally we have #(e^((ln(x))^2) )' = (2ln(x))/x*e^((ln(x))^2# Answer link Related questions What is the derivative of #y=3x^2e^(5x)# ? What is the derivative of #y=e^(3-2x)# ? What is the derivative of #f(theta)=e^(sin2theta)# ? What is the derivative of #f(x)=(e^(1/x))/x^2# ? What is the derivative of #f(x)=e^(pix)*cos(6x)# ? What is the derivative of #f(x)=x^4*e^sqrt(x)# ? What is the derivative of #f(x)=e^(-6x)+e# ? How do you find the derivative of #y=e^x#? How do you find the derivative of #y=e^(1/x)#? How do you find the derivative of #y=e^(2x)#? See all questions in Differentiating Exponential Functions with Base e Impact of this question 13803 views around the world You can reuse this answer Creative Commons License