How do you find the derivative of #h(theta)=2^(-theta)cospitheta#?

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]#

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