How do I find #f'(x)# for #f(x)=4^sqrt(x)# ? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Wataru Aug 30, 2014 By Chain Rule, we can find #f'(x)={(ln4)4^{sqrt{x}}}/{2sqrt{x}}#. Remember: #(b^x)'=(lnb)b^x# By Chain Rule, #f'(x)=(ln4)4^{sqrt{x}}cdot (sqrt{x})'=(ln4)4^{sqrt{x}}cdot{1}/{2sqrt{x}} ={(ln4)4^{sqrt{x}}}/{2sqrt{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)# ? 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))#? 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 5511 views around the world You can reuse this answer Creative Commons License