How do you differentiate #y=3^sinx#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Noah G Sep 24, 2016 #y = 3^sinx# #lny = ln(3^(sinx))# #lny = sinxln3# #1/y(dy/dx) = (cosx xx ln3 + 0 xx sinx)# #dy/dx = (ln3cosx)/(1/y)# #dy/dx = y xx ln3cosx# #dy/dx = 3^sinx xx ln3cosx# #dy/dx = 3^sinxln3cosx# Hopefully this helps! 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 16755 views around the world You can reuse this answer Creative Commons License