# Use part I of the Fundamental Theorem of Calculus to find the derivative of y= \int 9x \ to \...

## Question:

Use part I of the Fundamental Theorem of Calculus to find the derivative of {eq}y= \int 9x \ to \ cos(x) (cos(u^5)du) \frac {dy}{dx} = ? {/eq}

## A Derivative of an Integral:

To differentiate {eq}\int_{g(x)}^{h(x)} f(t) dt {/eq}, we apply the Fundamental Theorem of Calculus, {eq}\int_a^{g(x)} f(t) dt = f(g(x)) g'(x) {/eq} by, first, writing {eq}\int_{g(x)}^{h(x)} f(t) dt = -\_0^{g(x)} f(t) dt +?\_0^{h(x)} f(t) dt {/eq}.

The derivative of {eq}y= \int_{9x}^{\cos(x)} (cos(u^5) \, du) {/eq} is evaluated by applying Fundamental Theorem of Calculus, {eq}\dfrac{d}{dx} \int_a^{g(x)} f(t) \, dt = f(g(x)) \dfrac{d}{dx}g(x) {/eq}.

Then,

{eq}\begin{align*} \displaystyle y' &= \displaystyle \frac{d}{dx} \left(\int_{9x}^{\cos(x)} \cos(u^5) \, du \right) \\ &\\ &= \displaystyle \frac{d}{dx} \left(\int_{9x}^0 \cos(u^5) \, du + \int_{0}^{\cos(x)} \cos(u^5) \, du \right) \quad \text{[Apply integration properties, } \int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx \text{ such that all integrals have at least one bound of integral is a constant]} \\ &\\ &= \displaystyle \frac{d}{dx} \left( -\int_0^{9x} \cos(u^5) \, du + \int_{0}^{\cos(x)} (\cos(u^5) \, du \right) \quad \text{[Apply integration properties, } \int_a^b f(x) dx = -\int_b^a f(x) dx \text{ such that each the lower bound of all integrals is a constant]} \\ &\\ &= \displaystyle \frac{d}{dx} \left( -\int_0^{9x} \cos(u^5) \, du \right) + \frac{d}{dx} \left( \int_{0}^{\cos(x)} \cos(u^5) \, du \right) \\ &\\ &= \displaystyle -cos((9x)^5) \frac{d}{dx} \left( 9x \right) + \cos(\cos^5 x) \frac{d}{dx} \left( \cos(x) \right) \quad \text{[Apply Fundamental Theorem of Calculus]} \\ &\\ &= \displaystyle -cos(59049x^5) ( 9 ) + \cos(\cos^5 x) ( -\sin(x) ) \\ &\\ &= \displaystyle \color{blue}{ -9 \cos(59049x^5) - \sin (x) \cos(\cos^5 x) } \end{align*} {/eq}