How do you differentiate #y= 3^(2x+7)#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Jim H Mar 30, 2015 Use #d/(dx)(a^x)=a^x ln a# together with the chain rule: #d/(dx)(f(u))=f'(u) (du)/(dx)# For #y=3^(2x+7)# we get: #(dy)/(dx) = 3^(2x+7) ln3 (2) =2(3^(2x+7) ) ln3# Or use #2ln3=ln9# to write: #(dy)/(dx) = 3^(2x+7) ln9 # 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 1849 views around the world You can reuse this answer Creative Commons License