# logarithmic proof of quotient rule

###### Proof.

Let $f$ and $g$ be differentiable functions and $\displaystyle y=\frac{f(x)}{g(x)}$. Then $\ln y=\ln f(x)-\ln g(x)$. Thus, $\displaystyle\frac{1}{y}\cdot\frac{dy}{dx}=\frac{f^{\prime}(x)}{f(x)}-\frac{g^% {\prime}(x)}{g(x)}$. Therefore,

$\begin{array}[]{rl}\displaystyle\frac{dy}{dx}&\displaystyle=y\left(\frac{f^{% \prime}(x)}{f(x)}-\frac{g^{\prime}(x)}{g(x)}\right)\\ &\\ &\displaystyle=\frac{f(x)}{g(x)}\left(\frac{f^{\prime}(x)}{f(x)}-\frac{g^{% \prime}(x)}{g(x)}\right)\\ &\\ &\displaystyle=\frac{f^{\prime}(x)}{g(x)}-\frac{f(x)g^{\prime}(x)}{(g(x))^{2}}% \\ &\\ &\displaystyle=\frac{g(x)f^{\prime}(x)-f(x)g^{\prime}(x)}{(g(x))^{2}}.\end{array}$

Once students are familiar with the natural logarithm, the chain rule, and implicit differentiation, they typically have no problem following this proof of the quotient rule. Actually, with some prompting, they can produce a proof of the quotient rule to this one. This exercise is a great way for students to review many concepts from calculus  .

Title logarithmic proof of quotient rule LogarithmicProofOfQuotientRule 2013-03-22 16:18:51 2013-03-22 16:18:51 Wkbj79 (1863) Wkbj79 (1863) 7 Wkbj79 (1863) Proof msc 26A06 msc 97D40 LogarithmicProofOfProductRule