Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.5 Exercises - Page 362: 45

$h'(x)=-\frac{(\ln 2)(cos (\pi θ))+(\pi)(sin (\pi θ))}{2^θ}$

$h(θ)=2^{-θ}cos(\pi θ)$ $u=2^{-θ}$ $u’=-\ln 2 (2^{-θ})$ $v=cos (\pi θ)$ $v’=-\pi sin (\pi θ)$ $h’(x)=u’v+uv’$ $=(-\ln 2)(2^{-θ})(cos (\pi θ))+(2^{-θ})(-\pi)(sin (\pi θ))$ $=-\frac{(\ln 2)(cos (\pi θ))+(\pi)(sin (\pi θ))}{2^θ}$