Find $\dfrac{d}{dx}\int_{x^{3}}^{5}(t^{4}+1)^{1/3}\,dt$.

Solution To use Part 1 of the Fundamental Theorem of Calculus, the variable must be part of the upper limit of integration. So, we use the fact that $\int_{a}^{b}f(x)\,dx=-\int_{b}^{a}f(x)\,dx$ to interchange the limits of integration. $$\dfrac{d}{dx}\int_{x^{3}}^{5}(t^{4}+1)^{1/3}\,dt=\dfrac{d}{dx}\left[ {-} \int_{5}^{x^{3}}(t^{4}+1)^{1/3} dt\right] ={-}\dfrac{d}{dx} \int_{5}^{x^{3}}(t^{4}+1)^{1/3} dt$$

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Now we use the Chain Rule. We let $y=\int_{5}^{x^{3}}( t^{4}+1)^{1/3}dt$ and $u(x) =x^{3}.$

$$\begin{eqnarray*} \dfrac{d}{dx}\int_{x^{3}}^{5}(t^{4}+1)^{1/3}\,dt &=& -\dfrac{d}{dx}\int_{5}^{x^{3}}(t^{4}+1)^{1/3}\,dt=-\dfrac{dy}{dx} \underset{ \underset{ \color{#0066A7}{\hbox{Chain Rule}} } {\color{#0066A7}{\uparrow}} } {=} -\dfrac{dy}{du}\cdot\dfrac{du}{dx}\\ &=&-\dfrac{d}{du}\int_{5}^{u}( t^{4}+1) ^{1/3}dt\cdot \dfrac{du}{dx} \\ &=& -(u^{4}+1)^{1/3}\cdot \dfrac{du}{dx} {\color{#0066A7}\enspace{\color{#0066A7}{\hbox{Use the Fundamental Theorem}}}} \\ &=&-( x^{12}+1) ^{1/3}\cdot 3x^{2} \enspace{\color{#0066A7}{{\hbox{\({u=x}^{3}\); \(\dfrac{du}{dx}=3x^{2}\)}}}} \\ &=&-3x^{2}(x^{12}+1)^{1/3} \end{eqnarray*}$$