Use the quotient rule to find the derivative of $f(x)/g(x)$ where \begin{eqnarray} f(x)&=&5x^6-16x^5-4x^4\\ &+&72x^3-3x^2-2x+18 \end{eqnarray} and $g(x)=42x^2+32x+7$.

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Lesson Parent: 
Chapter 10 - The Quotient Rule
Problem: 

Use the quotient rule to find the derivative of $f(x)/g(x)$ where
\begin{eqnarray}
f(x)&=&5x^6-16x^5-4x^4\\
&+&72x^3-3x^2-2x+18
\end{eqnarray}
and
\[ g(x)=42x^2+32x+7\].

Answer: 

\begin{eqnarray} \frac{d}{dx}\left(\frac{f}{g}\right)&=&\left\{(42x^2+32x+7)(30x^5-80x^4-16x^3+216x^2-6x-2)\right.\\ &-&\left.(5x^6-16x^5-4x^4+72x^3-3x^2-2x+18)(84x+32)\right\}\\ &/&\left\{(42x^2+32x+7)^2\right\}. \end{eqnarray}

Solution: 

To use the quotient rule we first need to find $df/dx$ and $dg/dx$. We found the derivative of a polynomial here so we know that
\[
\frac{df}{dx}=30x^5-80x^4-16x^3+216x^2-6x-2
\]
and
\[
\frac{dg}{dx}=84x+32.
\]

Now the quotient rule says that
\[
\frac{d}{dx}\left(\frac{f}{g}\right)=\left(g\frac{df}{dx}-f\frac{dg}{dx}\right)/g^2.
\]
Plugging in our functions and derivatives we find that
\begin{eqnarray}
\frac{d}{dx}\left(\frac{f}{g}\right)&=&\left\{(42x^2+32x+7)(30x^5-80x^4-16x^3+216x^2-6x-2)\right.\\
&-&\left.(5x^6-16x^5-4x^4+72x^3-3x^2-2x+18)(84x+32)\right\}\\
&/&\left\{(42x^2+32x+7)^2\right\}.
\end{eqnarray}
That's it! We can stop here or simplify the derivative, but we have found the derivative that we wanted to find.