What is the derivative of #2^(cosx)#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer Cesareo R. Jan 10, 2017 #(dy)/(dx)=-log2sinx2^(cosx)# Explanation: Making #y=2^(cosx)#and applying #log# to both sides #log y = cosx log2# Deriving now #(dy)/y=-log2sinx dx# Rearranging #(dy)/(dx)=-log2ysinx=-log2sinx2^(cosx)# 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 10222 views around the world You can reuse this answer Creative Commons License