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Calculus
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Find the derivative of \(\displaystyle y=\sqrt{x^{3} - 13 x^{2} - 9 x + 4}\).

The outer function is \(\displaystyle f(u) = \sqrt{u}\) and the inner function \(\displaystyle u = x^{3} - 13 x^{2} - 9 x + 4\). The chain rule gives \(\displaystyle y'= \frac{df}{du}\frac{du}{dx}\). \(\displaystyle y' = \frac{1}{2 \sqrt{u}}(3 x^{2} - 26 x - 9) = \frac{\frac{3 x^{2}}{2} - 13 x - \frac{9}{2}}{\sqrt{x^{3} - 13 x^{2} - 9 x + 4}}\). Simplifying further gives \(\displaystyle y'= \frac{3 x^{2} - 26 x - 9}{2 \sqrt{x^{3} - 13 x^{2} - 9 x + 4}}\)

Download \(\LaTeX\) 

\begin{question}Find the derivative of $y=\sqrt{x^{3} - 13 x^{2} - 9 x + 4}$. 
    \soln{9cm}{The outer function is $f(u) = \sqrt{u}$ and the inner function $u = x^{3} - 13 x^{2} - 9 x + 4$. The chain rule gives $y'= \frac{df}{du}\frac{du}{dx}$. $y' = \frac{1}{2 \sqrt{u}}(3 x^{2} - 26 x - 9) = \frac{\frac{3 x^{2}}{2} - 13 x - \frac{9}{2}}{\sqrt{x^{3} - 13 x^{2} - 9 x + 4}}$. Simplifying further gives $y'= \frac{3 x^{2} - 26 x - 9}{2 \sqrt{x^{3} - 13 x^{2} - 9 x + 4}}$}

\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 y=\sqrt{x^{3} - 13 x^{2} - 9 x + 4} " src="/equation_images/%20%5Cdisplaystyle%20y%3D%5Csqrt%7Bx%5E%7B3%7D%20-%2013%20x%5E%7B2%7D%20-%209%20x%20%2B%204%7D%20" alt="LaTeX:  \displaystyle y=\sqrt{x^{3} - 13 x^{2} - 9 x + 4} " data-equation-content=" \displaystyle y=\sqrt{x^{3} - 13 x^{2} - 9 x + 4} " /> . </p> </p>
HTML for Canvas
<p> <p>The outer function is  <img class="equation_image" title=" \displaystyle f(u) = \sqrt{u} " src="/equation_images/%20%5Cdisplaystyle%20f%28u%29%20%3D%20%5Csqrt%7Bu%7D%20" alt="LaTeX:  \displaystyle f(u) = \sqrt{u} " data-equation-content=" \displaystyle f(u) = \sqrt{u} " />  and the inner function  <img class="equation_image" title=" \displaystyle u = x^{3} - 13 x^{2} - 9 x + 4 " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20x%5E%7B3%7D%20-%2013%20x%5E%7B2%7D%20-%209%20x%20%2B%204%20" alt="LaTeX:  \displaystyle u = x^{3} - 13 x^{2} - 9 x + 4 " data-equation-content=" \displaystyle u = x^{3} - 13 x^{2} - 9 x + 4 " /> . 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' = \frac{1}{2 \sqrt{u}}(3 x^{2} - 26 x - 9) = \frac{\frac{3 x^{2}}{2} - 13 x - \frac{9}{2}}{\sqrt{x^{3} - 13 x^{2} - 9 x + 4}} " src="/equation_images/%20%5Cdisplaystyle%20y%27%20%3D%20%5Cfrac%7B1%7D%7B2%20%5Csqrt%7Bu%7D%7D%283%20x%5E%7B2%7D%20-%2026%20x%20-%209%29%20%3D%20%5Cfrac%7B%5Cfrac%7B3%20x%5E%7B2%7D%7D%7B2%7D%20-%2013%20x%20-%20%5Cfrac%7B9%7D%7B2%7D%7D%7B%5Csqrt%7Bx%5E%7B3%7D%20-%2013%20x%5E%7B2%7D%20-%209%20x%20%2B%204%7D%7D%20" alt="LaTeX:  \displaystyle y' = \frac{1}{2 \sqrt{u}}(3 x^{2} - 26 x - 9) = \frac{\frac{3 x^{2}}{2} - 13 x - \frac{9}{2}}{\sqrt{x^{3} - 13 x^{2} - 9 x + 4}} " data-equation-content=" \displaystyle y' = \frac{1}{2 \sqrt{u}}(3 x^{2} - 26 x - 9) = \frac{\frac{3 x^{2}}{2} - 13 x - \frac{9}{2}}{\sqrt{x^{3} - 13 x^{2} - 9 x + 4}} " /> . Simplifying further gives  <img class="equation_image" title=" \displaystyle y'= \frac{3 x^{2} - 26 x - 9}{2 \sqrt{x^{3} - 13 x^{2} - 9 x + 4}} " src="/equation_images/%20%5Cdisplaystyle%20y%27%3D%20%5Cfrac%7B3%20x%5E%7B2%7D%20-%2026%20x%20-%209%7D%7B2%20%5Csqrt%7Bx%5E%7B3%7D%20-%2013%20x%5E%7B2%7D%20-%209%20x%20%2B%204%7D%7D%20" alt="LaTeX:  \displaystyle y'= \frac{3 x^{2} - 26 x - 9}{2 \sqrt{x^{3} - 13 x^{2} - 9 x + 4}} " data-equation-content=" \displaystyle y'= \frac{3 x^{2} - 26 x - 9}{2 \sqrt{x^{3} - 13 x^{2} - 9 x + 4}} " /> </p> </p>