How do you find the derivative of $h (θ) = 2^{- θ} cos π θ$ $h(theta)=2^(-theta)cospitheta$ ?

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Answer 1 · 2018-03-10T21:57:16

$- 2^{- θ} [(ln 2) cos π θ + π sin π θ]$ $-2^-theta[(ln2)cospitheta+pi sinpitheta]$

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:

$D[h(theta)]=D(2^-thetacospitheta)=[D(2^-theta)]*cospitheta+2^-thetaD(cospitheta)=$

$2^(-theta)*(ln2)(-1)(cospitheta)+ 2^-theta(-sinpitheta)(pi)=$

$-2^-theta[(ln2)cospitheta+pi sinpitheta]$

How do you find the derivative of h(theta)=2^(-theta)cospithetah(θ)=2−θcosπθh(theta)=2^(-theta)cospitheta?