How do you find the derivative of #y=sqrt(3^x-2^x)#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer A. S. Adikesavan Feb 19, 2017 #=1/(2y)(3^xln3-2^xln2 )# Explanation: Use #a^x=e^(lna x)# #y=(e^(ln3 x)-e^(ln2 x))^(1/2)# #y'=1/2(e^(ln3 x)-e^(ln2 x))^(-1/2)(e^(ln3 x)-e^(ln2 x))'# #=1/2(e^(ln3 x)-e^(ln2 x))^(-1/2)(ln3e^(ln3 x)-ln2e^(ln2 x))# #=1/(2y)(3^xln3-2^xln2 )# 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 1280 views around the world You can reuse this answer Creative Commons License