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Find the derivative of \(\displaystyle y = \frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}}\)
Taking the natural logarithm of both sides of the equation and expanding the right hand side gives: \begin{equation*}\ln(y) = \ln{\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right)} \end{equation*} Expanding the right hand side using the product and quotient properties of logarithms gives: \begin{equation*}\ln(y) = 7 \ln{\left(\sin{\left(x \right)} \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- x - 6 \ln{\left(2 - 8 x \right)} - 5 \ln{\left(3 x + 8 \right)} \end{equation*} Taking the derivative on both sides of the equation yields: \begin{equation*}\frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x} \end{equation*} Solving for \(\displaystyle y'\) and substituting out y using the original equation gives \begin{equation*}y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) \end{equation*} Using some Trigonometric identities to simplify gives \begin{equation*}y' = \left(- 4 \tan{\left(x \right)} + \frac{7}{\tan{\left(x \right)}}-1 - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) \end{equation*}
\begin{question}Find the derivative of $y = \frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}}$
\soln{9cm}{Taking the natural logarithm of both sides of the equation and expanding the right hand side gives:
\begin{equation*}\ln(y) = \ln{\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right)} \end{equation*}
Expanding the right hand side using the product and quotient properties of logarithms gives:
\begin{equation*}\ln(y) = 7 \ln{\left(\sin{\left(x \right)} \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- x - 6 \ln{\left(2 - 8 x \right)} - 5 \ln{\left(3 x + 8 \right)} \end{equation*}
Taking the derivative on both sides of the equation yields:
\begin{equation*}\frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x} \end{equation*}
Solving for $y'$ and substituting out y using the original equation gives
\begin{equation*}y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) \end{equation*}
Using some Trigonometric identities to simplify gives
\begin{equation*}y' = \left(- 4 \tan{\left(x \right)} + \frac{7}{\tan{\left(x \right)}}-1 - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) \end{equation*}
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\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 = \frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} " src="/equation_images/%20%5Cdisplaystyle%20y%20%3D%20%5Cfrac%7Be%5E%7B-%20x%7D%20%5Csin%5E%7B7%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%5E%7B4%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Cleft%282%20-%208%20x%5Cright%29%5E%7B6%7D%20%5Cleft%283%20x%20%2B%208%5Cright%29%5E%7B5%7D%7D%20" alt="LaTeX: \displaystyle y = \frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} " data-equation-content=" \displaystyle y = \frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} " /> </p> </p><p> <p>Taking the natural logarithm of both sides of the equation and expanding the right hand side gives:
<img class="equation_image" title=" \ln(y) = \ln{\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right)} " src="/equation_images/%20%5Cln%28y%29%20%3D%20%5Cln%7B%5Cleft%28%5Cfrac%7Be%5E%7B-%20x%7D%20%5Csin%5E%7B7%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%5E%7B4%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Cleft%282%20-%208%20x%5Cright%29%5E%7B6%7D%20%5Cleft%283%20x%20%2B%208%5Cright%29%5E%7B5%7D%7D%20%5Cright%29%7D%20%20%20" alt="LaTeX: \ln(y) = \ln{\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right)} " data-equation-content=" \ln(y) = \ln{\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right)} " />
Expanding the right hand side using the product and quotient properties of logarithms gives:
<img class="equation_image" title=" \ln(y) = 7 \ln{\left(\sin{\left(x \right)} \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- x - 6 \ln{\left(2 - 8 x \right)} - 5 \ln{\left(3 x + 8 \right)} " src="/equation_images/%20%5Cln%28y%29%20%3D%207%20%5Cln%7B%5Cleft%28%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D%20%2B%204%20%5Cln%7B%5Cleft%28%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D-%20x%20-%206%20%5Cln%7B%5Cleft%282%20-%208%20x%20%5Cright%29%7D%20-%205%20%5Cln%7B%5Cleft%283%20x%20%2B%208%20%5Cright%29%7D%20%20%20" alt="LaTeX: \ln(y) = 7 \ln{\left(\sin{\left(x \right)} \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- x - 6 \ln{\left(2 - 8 x \right)} - 5 \ln{\left(3 x + 8 \right)} " data-equation-content=" \ln(y) = 7 \ln{\left(\sin{\left(x \right)} \right)} + 4 \ln{\left(\cos{\left(x \right)} \right)}- x - 6 \ln{\left(2 - 8 x \right)} - 5 \ln{\left(3 x + 8 \right)} " />
Taking the derivative on both sides of the equation yields:
<img class="equation_image" title=" \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x} " src="/equation_images/%20%5Cfrac%7By%27%7D%7By%7D%20%3D%20-%20%5Cfrac%7B4%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20-%201%20%2B%20%5Cfrac%7B7%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%20-%20%5Cfrac%7B15%7D%7B3%20x%20%2B%208%7D%20%2B%20%5Cfrac%7B48%7D%7B2%20-%208%20x%7D%20%20%20" alt="LaTeX: \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x} " data-equation-content=" \frac{y'}{y} = - \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x} " />
Solving for <img class="equation_image" title=" \displaystyle y' " src="/equation_images/%20%5Cdisplaystyle%20y%27%20" alt="LaTeX: \displaystyle y' " data-equation-content=" \displaystyle y' " /> and substituting out y using the original equation gives
<img class="equation_image" title=" y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) " src="/equation_images/%20y%27%20%3D%20%5Cleft%28-%20%5Cfrac%7B4%20%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%20-%201%20%2B%20%5Cfrac%7B7%20%5Ccos%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Csin%7B%5Cleft%28x%20%5Cright%29%7D%7D%20-%20%5Cfrac%7B15%7D%7B3%20x%20%2B%208%7D%20%2B%20%5Cfrac%7B48%7D%7B2%20-%208%20x%7D%5Cright%29%5Cleft%28%5Cfrac%7Be%5E%7B-%20x%7D%20%5Csin%5E%7B7%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%5E%7B4%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Cleft%282%20-%208%20x%5Cright%29%5E%7B6%7D%20%5Cleft%283%20x%20%2B%208%5Cright%29%5E%7B5%7D%7D%20%5Cright%29%20%20%20" alt="LaTeX: y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) " data-equation-content=" y' = \left(- \frac{4 \sin{\left(x \right)}}{\cos{\left(x \right)}} - 1 + \frac{7 \cos{\left(x \right)}}{\sin{\left(x \right)}} - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) " />
Using some Trigonometric identities to simplify gives
<img class="equation_image" title=" y' = \left(- 4 \tan{\left(x \right)} + \frac{7}{\tan{\left(x \right)}}-1 - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) " src="/equation_images/%20y%27%20%3D%20%5Cleft%28-%204%20%5Ctan%7B%5Cleft%28x%20%5Cright%29%7D%20%2B%20%5Cfrac%7B7%7D%7B%5Ctan%7B%5Cleft%28x%20%5Cright%29%7D%7D-1%20-%20%5Cfrac%7B15%7D%7B3%20x%20%2B%208%7D%20%2B%20%5Cfrac%7B48%7D%7B2%20-%208%20x%7D%5Cright%29%5Cleft%28%5Cfrac%7Be%5E%7B-%20x%7D%20%5Csin%5E%7B7%7D%7B%5Cleft%28x%20%5Cright%29%7D%20%5Ccos%5E%7B4%7D%7B%5Cleft%28x%20%5Cright%29%7D%7D%7B%5Cleft%282%20-%208%20x%5Cright%29%5E%7B6%7D%20%5Cleft%283%20x%20%2B%208%5Cright%29%5E%7B5%7D%7D%20%5Cright%29%20%20%20" alt="LaTeX: y' = \left(- 4 \tan{\left(x \right)} + \frac{7}{\tan{\left(x \right)}}-1 - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) " data-equation-content=" y' = \left(- 4 \tan{\left(x \right)} + \frac{7}{\tan{\left(x \right)}}-1 - \frac{15}{3 x + 8} + \frac{48}{2 - 8 x}\right)\left(\frac{e^{- x} \sin^{7}{\left(x \right)} \cos^{4}{\left(x \right)}}{\left(2 - 8 x\right)^{6} \left(3 x + 8\right)^{5}} \right) " />
</p> </p>