## Calculus 8th Edition

$h'(x) = -\frac{sin^4(\frac{1}{x})}{x^2}$
We are told to find the derivative using Part 1 of the Fundamental Theorem of Calculus given the expression: $h(x) = \int ^{\frac{1}{x}}_{2}sin^4(t )dt$ Since the upper bound is a function we need to use the chain rule: $u = \frac{1}{x}$ $u' = -\frac{1}{x^2}$ Since $h(x) = \int^{b}_{a}f(t)dt$ $h'(x) = f(x)$ $h'(u) = sin^4(u)u'$ $h'(x) = -sin^4(\frac{1}{x}) \frac{1}{x^2}$ $h'(x) = -\frac{sin^4(\frac{1}{x})}{x^2}$