Alternate Derivative Proofs

In this note, I'm going to demonstrate another way to prove the derivative formulas $\frac{d}{dx}f(x)\times g(x)=f'(x)g(x)+g'(x)f(x)$ and $\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}$ using logarithmic differentiation. The most commonly taught method is with the definition of a derivative, but this can get ugly and can easily result in a stupid mistake.

First, let's take a look at the derivative of the natural logarithm: $\frac{d}{dx}\ln x.$

If $y=\ln x,$ then $e^y=x.$ Derive both sides with respect to $x.$ $\frac{d}{dy}e^y\frac{dy}{dx}=\frac{d}{dx}x$ $e^y\frac{dy}{dx}=1$ $\frac{dy}{dx}=\frac{1}{e^y}$ $\frac{dy}{dx}=\frac{1}{e^{\ln x}}=\frac{1}{x}$ So that is the derivative of $\ln x.$ If $u$ is a differentiable function in $x,$ then by the chain rule, $\frac{d}{dx}\ln u=\frac{u'}{u}.$ Now we can move to logarithmic differentiation.

Logarithmic differentiation is commonly used for really ugly functions to differentiate. You use it if you are told to derive $\frac{(x-2)^4(x+1)^3}{(x-3)^6},$ for example. If you try to use the chain rule, product rule, and quotient rule for this, you are in for one heck of a ride, and the answer you come up with has a really good chance of being wrong. You can take the natural logarithm of both sides and apply the properties of logarithms and the chain rule to find the answer.

$y=\dfrac{(x-2)^4(x+1)^3}{(x-3)^6}$ $\ln y=\ln \dfrac{(x-2)^4(x+1)^3}{(x-3)^6}$ $\ln y=4\ln(x-2)+3\ln(x+1)-6\ln(x-3)$ $\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{4}{x-2}+\dfrac{3}{x+1}-\dfrac{6}{x-3}$ $\dfrac{dy}{dx}=y\left(\dfrac{4}{x-2}+\dfrac{3}{x+1}-\dfrac{6}{x-3}\right)$ $\dfrac{dy}{dx}=\left(\dfrac{(x-2)^4(x+1)^3}{(x-3)^6}\right)\left(\dfrac{4}{x-2}+\dfrac{3}{x+1}-\dfrac{6}{x-3}\right)$ This is much easier to evaluate for values of $x.$ If you are actually trying to expand an expression like this, then you are being trolled. A time where you might expand the expression is if you are deriving $\frac{(x+1)^4}{(x-2)^3}.$

So now onto the proofs. $f,$ $g,$ and $F$ are differentiable functions of $x.$

Let's start with the product rule. Let $F=fg$ $F=fg$ $\ln F=\ln fg$ $\ln F=\ln f+\ln g$ $\dfrac{F'}{F}=\dfrac{f'}{f}+\dfrac{g'}{g}$ $\dfrac{F'}{F}=\dfrac{f'g+g'f}{fg}$ $F'=F\left(\dfrac{f'g+g'f}{fg}\right)$ $F'=f'g+g'f$ So this is the proof of the product rule. Now let's move on the quotient rule. Now let $F=\dfrac{f}{g}.$ $F=\dfrac{f}{g}$ $\ln F=\ln\dfrac{f}{g}$ $\ln F=\ln f-\ln g$ $\dfrac{F'}{F}=\dfrac{f'}{f}-\dfrac{g'}{g}$ $\dfrac{F'}{F}=\dfrac{f'g-g'f}{fg}$ $F'=F\left(\dfrac{f'g-g'f}{fg}\right)$ $F'=\left(\dfrac{f}{g}\right)\left(\dfrac{f'g-g'f}{fg}\right)$ $F'=\dfrac{f'g-g'f}{g^2}$ And now the quotient rule has been proven.

In my opinion, this is a lot easier than trying to figure out what you need to add and subtract to the expression found using the definition of a derivative. Hopefully you do too! Thanks for reading this post, and remember to occasionally check #TrevorsTips for tricks to solve problems.