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Find the derivative of \(\displaystyle F(x) = \int\limits_{9 x - 8}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt\).

Before using the Fundamental Theorem of Calculus part I the integral must be split into two integrals and the limits of integration must be reversed on the 2nd integral.
This gives \(\displaystyle F(x) = \int\limits_{0}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt-\int\limits_{0}^{9 x - 8} \left(9 t - 8\right)\, dt\). Now using the Fundamental Theorem of Calculus part I and the chain rule with \(\displaystyle u=\cos{\left(x \right)}\) and \(\displaystyle F(u)=\int\limits_{0}^{u} \left(9 t - 8\right)\, dt\) and \(\displaystyle v=9 x - 8\) and \(\displaystyle F(v)=- \int\limits_{0}^{v} \left(9 t - 8\right)\, dt\) gives: \(\displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(9 u - 8\right)\left(- \sin{\left(x \right)}\right)+\left(8 - 9 v\right)\left(9\right)=- 729 x - \left(9 \cos{\left(x \right)} - 8\right) \sin{\left(x \right)} + 720\)

Download \(\LaTeX\) 

\begin{question}Find the derivative of $F(x) = \int\limits_{9 x - 8}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt$. 
    \soln{9cm}{Before using the Fundamental Theorem of Calculus part I the integral must be split into two integrals and the limits of integration must be reversed on the 2nd integral.\newline This gives $F(x) = \int\limits_{0}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt-\int\limits_{0}^{9 x - 8} \left(9 t - 8\right)\, dt$. Now using the Fundamental Theorem of Calculus part I and the chain rule with $u=\cos{\left(x \right)}$  and $F(u)=\int\limits_{0}^{u} \left(9 t - 8\right)\, dt$ and $v=9 x - 8$  and $F(v)=- \int\limits_{0}^{v} \left(9 t - 8\right)\, dt$ gives: $F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(9 u - 8\right)\left(- \sin{\left(x \right)}\right)+\left(8 - 9 v\right)\left(9\right)=- 729 x - \left(9 \cos{\left(x \right)} - 8\right) \sin{\left(x \right)} + 720$}

\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_{9 x - 8}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B9%20x%20-%208%7D%5E%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cleft%289%20t%20-%208%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = \int\limits_{9 x - 8}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{9 x - 8}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt " /> . </p> </p>
HTML for Canvas
<p> <p>Before using the Fundamental Theorem of Calculus part I the integral must be split into two integrals and the limits of integration must be reversed on the 2nd integral.<br> This gives  <img class="equation_image" title=" \displaystyle F(x) = \int\limits_{0}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt-\int\limits_{0}^{9 x - 8} \left(9 t - 8\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28x%29%20%3D%20%5Cint%5Climits_%7B0%7D%5E%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20%5Cleft%289%20t%20-%208%5Cright%29%5C%2C%20dt-%5Cint%5Climits_%7B0%7D%5E%7B9%20x%20-%208%7D%20%5Cleft%289%20t%20-%208%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(x) = \int\limits_{0}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt-\int\limits_{0}^{9 x - 8} \left(9 t - 8\right)\, dt " data-equation-content=" \displaystyle F(x) = \int\limits_{0}^{\cos{\left(x \right)}} \left(9 t - 8\right)\, dt-\int\limits_{0}^{9 x - 8} \left(9 t - 8\right)\, dt " /> . Now 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_{0}^{u} \left(9 t - 8\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28u%29%3D%5Cint%5Climits_%7B0%7D%5E%7Bu%7D%20%5Cleft%289%20t%20-%208%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(u)=\int\limits_{0}^{u} \left(9 t - 8\right)\, dt " data-equation-content=" \displaystyle F(u)=\int\limits_{0}^{u} \left(9 t - 8\right)\, dt " />  and  <img class="equation_image" title=" \displaystyle v=9 x - 8 " src="/equation_images/%20%5Cdisplaystyle%20v%3D9%20x%20-%208%20" alt="LaTeX:  \displaystyle v=9 x - 8 " data-equation-content=" \displaystyle v=9 x - 8 " />   and  <img class="equation_image" title=" \displaystyle F(v)=- \int\limits_{0}^{v} \left(9 t - 8\right)\, dt " src="/equation_images/%20%5Cdisplaystyle%20F%28v%29%3D-%20%5Cint%5Climits_%7B0%7D%5E%7Bv%7D%20%5Cleft%289%20t%20-%208%5Cright%29%5C%2C%20dt%20" alt="LaTeX:  \displaystyle F(v)=- \int\limits_{0}^{v} \left(9 t - 8\right)\, dt " data-equation-content=" \displaystyle F(v)=- \int\limits_{0}^{v} \left(9 t - 8\right)\, dt " />  gives:  <img class="equation_image" title=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(9 u - 8\right)\left(- \sin{\left(x \right)}\right)+\left(8 - 9 v\right)\left(9\right)=- 729 x - \left(9 \cos{\left(x \right)} - 8\right) \sin{\left(x \right)} + 720 " src="/equation_images/%20%5Cdisplaystyle%20F%27%28x%29%3D%5Cfrac%7BdF%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20%2B%20%5Cfrac%7BdF%7D%7Bdv%7D%5Cfrac%7Bdv%7D%7Bdx%7D%3D%20%5Cleft%289%20u%20-%208%5Cright%29%5Cleft%28-%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%5Cright%29%2B%5Cleft%288%20-%209%20v%5Cright%29%5Cleft%289%5Cright%29%3D-%20729%20x%20-%20%5Cleft%289%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20-%208%5Cright%29%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20720%20" alt="LaTeX:  \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(9 u - 8\right)\left(- \sin{\left(x \right)}\right)+\left(8 - 9 v\right)\left(9\right)=- 729 x - \left(9 \cos{\left(x \right)} - 8\right) \sin{\left(x \right)} + 720 " data-equation-content=" \displaystyle F'(x)=\frac{dF}{du}\frac{du}{dx} + \frac{dF}{dv}\frac{dv}{dx}= \left(9 u - 8\right)\left(- \sin{\left(x \right)}\right)+\left(8 - 9 v\right)\left(9\right)=- 729 x - \left(9 \cos{\left(x \right)} - 8\right) \sin{\left(x \right)} + 720 " /> </p> </p>