How do you differentiate #y=7^(x^2)#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer 256 Feb 5, 2017 #y=7^(x^2) => y'=(2ln(7)x)7^(x^2)# Explanation: Since #forall r in RR, x^r=e^(rln(x))# It is true that #7^(x^2)=e^(x^2ln(7))# Then by the chain rule #(f(g(x)))'=f'(g(x))g'(x)# #=> (e^(x^2ln(7)))'=e^(x^2ln(7))2ln(7)x=7^(x^2)(2ln(7)x)# Answer link Related questions How do I find #f'(x)# for #f(x)=5^x# ? How do I find #f'(x)# for #f(x)=3^-x# ? How do I find #f'(x)# for #f(x)=x^2*10^(2x)# ? How do I find #f'(x)# for #f(x)=4^sqrt(x)# ? What is the derivative of #f(x)=b^x# ? What is the derivative of 10^x? How do you find the derivative of #x^(2x)#? How do you find the derivative of #f(x)=pi^cosx#? How do you find the derivative of #y=(sinx)^(x^3)#? How do you find the derivative of #y=ln(1+e^(2x))#? See all questions in Differentiating Exponential Functions with Other Bases Impact of this question 6904 views around the world You can reuse this answer Creative Commons License