How do you find the derivative of #h(theta)=2^(-theta)cospitheta#? Calculus Differentiating Exponential Functions Differentiating Exponential Functions with Other Bases 1 Answer dhilak Mar 10, 2018 #-2^-theta[(ln2)cospitheta+pi sinpitheta]# Explanation: Using the formula: #D[f(x)*g(x)]=f'(x)*g(x)+f(x)g'(x)# and #D(a^f(x))=a^f(x)(lna)[f'(x)]# and #Dcos[g(x)]=-{sin[g(x)]}*g'(x)# could get to: #D[h(theta)]=D(2^-thetacospitheta)=[D(2^-theta)]*cospitheta+2^-thetaD(cospitheta)=# #2^(-theta)*(ln2)(-1)(cospitheta)+ 2^-theta(-sinpitheta)(pi)=# #-2^-theta[(ln2)cospitheta+pi sinpitheta]# 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 2225 views around the world You can reuse this answer Creative Commons License