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Questions: Algebra BusinessCalculus
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Find the derivative of \(\displaystyle y=\tan{\left(x^{2} + 11 x - 13 \right)}\).
The outer function is \(\displaystyle f(u) = \tan{\left(u \right)}\) and the inner function \(\displaystyle u = x^{2} + 11 x - 13\). The chain rule gives \(\displaystyle y'= \frac{df}{du}\frac{du}{dx}\). \(\displaystyle y' = \tan^{2}{\left(u \right)} + 1(2 x + 11) = \left(2 x + 11\right) \left(\tan^{2}{\left(x^{2} + 11 x - 13 \right)} + 1\right)\).
\begin{question}Find the derivative of $y=\tan{\left(x^{2} + 11 x - 13 \right)}$.
\soln{9cm}{The outer function is $f(u) = \tan{\left(u \right)}$ and the inner function $u = x^{2} + 11 x - 13$. The chain rule gives $y'= \frac{df}{du}\frac{du}{dx}$. $y' = \tan^{2}{\left(u \right)} + 1(2 x + 11) = \left(2 x + 11\right) \left(\tan^{2}{\left(x^{2} + 11 x - 13 \right)} + 1\right)$. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Find the derivative of <img class="equation_image" title=" \displaystyle y=\tan{\left(x^{2} + 11 x - 13 \right)} " src="/equation_images/%20%5Cdisplaystyle%20y%3D%5Ctan%7B%5Cleft%28x%5E%7B2%7D%20%2B%2011%20x%20-%2013%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle y=\tan{\left(x^{2} + 11 x - 13 \right)} " data-equation-content=" \displaystyle y=\tan{\left(x^{2} + 11 x - 13 \right)} " /> . </p> </p><p> <p>The outer function is <img class="equation_image" title=" \displaystyle f(u) = \tan{\left(u \right)} " src="/equation_images/%20%5Cdisplaystyle%20f%28u%29%20%3D%20%5Ctan%7B%5Cleft%28u%20%5Cright%29%7D%20" alt="LaTeX: \displaystyle f(u) = \tan{\left(u \right)} " data-equation-content=" \displaystyle f(u) = \tan{\left(u \right)} " /> and the inner function <img class="equation_image" title=" \displaystyle u = x^{2} + 11 x - 13 " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20x%5E%7B2%7D%20%2B%2011%20x%20-%2013%20" alt="LaTeX: \displaystyle u = x^{2} + 11 x - 13 " data-equation-content=" \displaystyle u = x^{2} + 11 x - 13 " /> . The chain rule gives <img class="equation_image" title=" \displaystyle y'= \frac{df}{du}\frac{du}{dx} " src="/equation_images/%20%5Cdisplaystyle%20y%27%3D%20%5Cfrac%7Bdf%7D%7Bdu%7D%5Cfrac%7Bdu%7D%7Bdx%7D%20" alt="LaTeX: \displaystyle y'= \frac{df}{du}\frac{du}{dx} " data-equation-content=" \displaystyle y'= \frac{df}{du}\frac{du}{dx} " /> . <img class="equation_image" title=" \displaystyle y' = \tan^{2}{\left(u \right)} + 1(2 x + 11) = \left(2 x + 11\right) \left(\tan^{2}{\left(x^{2} + 11 x - 13 \right)} + 1\right) " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Ctan%5E%7B2%7D%7B%5Cleft%28u%20%5Cright%29%7D%20%2B%201%282%20x%20%2B%2011%29%20%3D%20%5Cleft%282%20x%20%2B%2011%5Cright%29%20%5Cleft%28%5Ctan%5E%7B2%7D%7B%5Cleft%28x%5E%7B2%7D%20%2B%2011%20x%20-%2013%20%5Cright%29%7D%20%2B%201%5Cright%29%20" alt="LaTeX: \displaystyle y' = \tan^{2}{\left(u \right)} + 1(2 x + 11) = \left(2 x + 11\right) \left(\tan^{2}{\left(x^{2} + 11 x - 13 \right)} + 1\right) " data-equation-content=" \displaystyle y' = \tan^{2}{\left(u \right)} + 1(2 x + 11) = \left(2 x + 11\right) \left(\tan^{2}{\left(x^{2} + 11 x - 13 \right)} + 1\right) " /> . </p> </p>