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Calculus
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Find the derivative of \(\displaystyle F(x) = \int\limits_{-5}^{\frac{1}{x^{2}}} \left(6 t^{2} + 2 t - 6\right)\, dt\).

Using the Fundamental Theorem of Calculus part I and the chain rule with \(\displaystyle u=\frac{1}{x^{2}}\) and \(\displaystyle F(u)=\int\limits_{-5}^{u} \left(6 t^{2} + 2 t - 6\right)\, dt\) gives: \(\displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(6 u^{2} + 2 u - 6\right)\left(- \frac{2}{x^{3}}\right)=- \frac{2 \left(-6 + \frac{2}{x^{2}} + \frac{6}{x^{4}}\right)}{x^{3}}\)

Download \(\LaTeX\) 

\begin{question}Find the derivative of $F(x) = \int\limits_{-5}^{\frac{1}{x^{2}}} \left(6 t^{2} + 2 t - 6\right)\, dt$. 
    \soln{9cm}{Using the Fundamental Theorem of Calculus part I and the chain rule with $u=\frac{1}{x^{2}}$  and $F(u)=\int\limits_{-5}^{u} \left(6 t^{2} + 2 t - 6\right)\, dt$ gives: $F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(6 u^{2} + 2 u - 6\right)\left(- \frac{2}{x^{3}}\right)=- \frac{2 \left(-6 + \frac{2}{x^{2}} + \frac{6}{x^{4}}\right)}{x^{3}}$}

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Find the derivative of  <img class="equation_image" title=" \displaystyle F(x) = \int\limits_{-5}^{\frac{1}{x^{2}}} \left(6 t^{2} + 2 t - 6\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B-5%7D%5E%7B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%7D%7D%20%5Cleft%286%20t%5E%7B2%7D%20%2B%202%20t%20-%206%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = \int\limits_{-5}^{\frac{1}{x^{2}}} \left(6 t^{2} + 2 t - 6\right)\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{-5}^{\frac{1}{x^{2}}} \left(6 t^{2} + 2 t - 6\right)\, dt " /> . </p> </p>
HTML for Canvas
<p> <p>Using the Fundamental Theorem of Calculus part I and the chain rule with  <img class="equation_image" title=" \displaystyle u=\frac{1}{x^{2}} " src="/equation_images/%20%5Cdisplaystyle%20u%3D%5Cfrac%7B1%7D%7Bx%5E%7B2%7D%7D%20" alt="LaTeX:  \displaystyle u=\frac{1}{x^{2}} " data-equation-content=" \displaystyle u=\frac{1}{x^{2}} " />   and  <img class="equation_image" title=" \displaystyle F(u)=\int\limits_{-5}^{u} \left(6 t^{2} + 2 t - 6\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28u%29%3D%5Cint%5Climits_%7B-5%7D%5E%7Bu%7D%20%5Cleft%286%20t%5E%7B2%7D%20%2B%202%20t%20-%206%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(u)=\int\limits_{-5}^{u} \left(6 t^{2} + 2 t - 6\right)\, dt " data-equation-content=" \displaystyle F(u)=\int\limits_{-5}^{u} \left(6 t^{2} + 2 t - 6\right)\, dt " />  gives:  <img class="equation_image" title=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(6 u^{2} + 2 u - 6\right)\left(- \frac{2}{x^{3}}\right)=- \frac{2 \left(-6 + \frac{2}{x^{2}} + \frac{6}{x^{4}}\right)}{x^{3}} " src="/equation_images/%20%5Cdisplaystyle%20F%27%28x%29%3D%5Cfrac%7BdF%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20%3D%20%5Cleft%286%20u%5E%7B2%7D%20%2B%202%20u%20-%206%5Cright%29%5Cleft%28-%20%5Cfrac%7B2%7D%7Bx%5E%7B3%7D%7D%5Cright%29%3D-%20%5Cfrac%7B2%20%5Cleft%28-6%20%2B%20%5Cfrac%7B2%7D%7Bx%5E%7B2%7D%7D%20%2B%20%5Cfrac%7B6%7D%7Bx%5E%7B4%7D%7D%5Cright%29%7D%7Bx%5E%7B3%7D%7D%20" alt="LaTeX:  \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(6 u^{2} + 2 u - 6\right)\left(- \frac{2}{x^{3}}\right)=- \frac{2 \left(-6 + \frac{2}{x^{2}} + \frac{6}{x^{4}}\right)}{x^{3}} " data-equation-content=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(6 u^{2} + 2 u - 6\right)\left(- \frac{2}{x^{3}}\right)=- \frac{2 \left(-6 + \frac{2}{x^{2}} + \frac{6}{x^{4}}\right)}{x^{3}} " /> </p> </p>