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

Mar 10, 2018

$- {2}^{-} \theta \left[\left(\ln 2\right) \cos \pi \theta + \pi \sin \pi \theta\right]$

#### Explanation:

Using the formula: $D \left[f \left(x\right) \cdot g \left(x\right)\right] = f ' \left(x\right) \cdot g \left(x\right) + f \left(x\right) g ' \left(x\right)$ and

$D \left({a}^{f} \left(x\right)\right) = {a}^{f} \left(x\right) \left(\ln a\right) \left[f ' \left(x\right)\right]$ and $D \cos \left[g \left(x\right)\right] = - \left\{\sin \left[g \left(x\right)\right]\right\} \cdot g ' \left(x\right)$

could get to:

$D \left[h \left(\theta\right)\right] = D \left({2}^{-} \theta \cos \pi \theta\right) = \left[D \left({2}^{-} \theta\right)\right] \cdot \cos \pi \theta + {2}^{-} \theta D \left(\cos \pi \theta\right) =$

${2}^{- \theta} \cdot \left(\ln 2\right) \left(- 1\right) \left(\cos \pi \theta\right) + {2}^{-} \theta \left(- \sin \pi \theta\right) \left(\pi\right) =$

$- {2}^{-} \theta \left[\left(\ln 2\right) \cos \pi \theta + \pi \sin \pi \theta\right]$