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

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

Download \(\LaTeX\) 

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

\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_{2}^{\cos{\left(x \right)}} \left(2 t + 6\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B2%7D%5E%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cleft%282%20t%20%2B%206%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = \int\limits_{2}^{\cos{\left(x \right)}} \left(2 t + 6\right)\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{2}^{\cos{\left(x \right)}} \left(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=\cos{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20u%3D%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle u=\cos{\left(x \right)} " data-equation-content=" \displaystyle u=\cos{\left(x \right)} " />   and  <img class="equation_image" title=" \displaystyle F(u)=\int\limits_{2}^{u} \left(2 t + 6\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28u%29%3D%5Cint%5Climits_%7B2%7D%5E%7Bu%7D%20%5Cleft%282%20t%20%2B%206%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(u)=\int\limits_{2}^{u} \left(2 t + 6\right)\, dt " data-equation-content=" \displaystyle F(u)=\int\limits_{2}^{u} \left(2 t + 6\right)\, dt " />  gives:  <img class="equation_image" title=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(2 u + 6\right)\left(- \sin{\left(x \right)}\right)=- \left(2 \cos{\left(x \right)} + 6\right) \sin{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20F%27%28x%29%3D%5Cfrac%7BdF%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20%3D%20%5Cleft%282%20u%20%2B%206%5Cright%29%5Cleft%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%5Cright%29%3D-%20%5Cleft%282%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%206%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(2 u + 6\right)\left(- \sin{\left(x \right)}\right)=- \left(2 \cos{\left(x \right)} + 6\right) \sin{\left(x \right)} " data-equation-content=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} = \left(2 u + 6\right)\left(- \sin{\left(x \right)}\right)=- \left(2 \cos{\left(x \right)} + 6\right) \sin{\left(x \right)} " /> </p> </p>