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# 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.

Note by Trevor B.
2 years, 7 months ago

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Can you use similar techniques to prove that $$\frac{d}{dx}x^n=nx^{n-1}?$$ · 2 years, 7 months ago

Let $$y = x^n$$. Taking the natural logarithm of both sides gives $$\ln y = n \ln x$$. Differentiating, one has $\frac{d}{dy} \cdot \frac{dy}{dx} \ln y = n \frac{d}{dx} \ln x$ which simplifies to $$\frac{dy}{dx}\frac{1}{y} = \frac{n}{x}$$. Substituting $$y = x^{n}$$ back into the equation and rearranging, we find that $$\frac{dy}{dx} = nx^{n-1}$$, as desired. · 2 years, 7 months ago

Good job! · 2 years, 7 months ago

Thanks! It took me a bit too long to format it, though... · 2 years, 7 months ago

Yup, we can.

Let's say $$y = x^n$$

$$\frac{d}{dx} y = \frac{d}{dx} x^n$$

Take the natural log of both sides, and using logarithmic properties we get...

$$\frac{d}{dx} lny = \frac{d}{dx} ln x^n = \frac{d}{dx} nln x$$

$$\frac{1}{y} \frac{dy}{dx} = n\frac{1}{x}$$

Multiply both sides by $$y$$.

$$y\frac{1}{y} \frac{dy}{dx} = ny\frac{1}{x}$$

And so we get...

$$\frac{dy}{dx} = ny\frac{1}{x}$$ and remember, $$y = x^n$$, hence substituting it we get,

$$\frac{dy}{dx} = nx^n\frac{1}{x}$$

Using exponent properties, I can say...

$$\frac{dy}{dx} = nx^n x^{-1}$$

$$\frac{dy}{dx} = nx^{n-1}$$

$$y = x^n \Rightarrow \frac{d}{dx} x^n = nx^{n-1}$$

I always used to think of interesting ways and they connect beautifully, it's always nice to see something solved in more than one way. ^^ · 2 years, 7 months ago

Good job! · 2 years, 7 months ago

Can anyone think of what else you can use logarithmic differentiation for (in terms of proofs)? · 2 years, 7 months ago

Something like $$x^x$$ or even $$x^{x^x}$$ · 2 years, 7 months ago

right........ it is especially helpful when a function is raised to another function. Here is an interesting example find the derivative of y=x^x^x^x..... up to infinity (try to find the derivative) · 2 years, 4 months ago

Luv u de · 1 year ago