Alternate Derivative Proofs

In this note, I'm going to demonstrate another way to prove the derivative formulas ddxf(x)×g(x)=f(x)g(x)+g(x)f(x)\frac{d}{dx}f(x)\times g(x)=f'(x)g(x)+g'(x)f(x) and ddxf(x)g(x)=f(x)g(x)g(x)f(x)(g(x))2\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: ddxlnx.\frac{d}{dx}\ln x.

If y=lnx,y=\ln x, then ey=x.e^y=x. Derive both sides with respect to x.x. ddyeydydx=ddxx\frac{d}{dy}e^y\frac{dy}{dx}=\frac{d}{dx}x eydydx=1e^y\frac{dy}{dx}=1 dydx=1ey\frac{dy}{dx}=\frac{1}{e^y} dydx=1elnx=1x\frac{dy}{dx}=\frac{1}{e^{\ln x}}=\frac{1}{x} So that is the derivative of lnx.\ln x. If uu is a differentiable function in x,x, then by the chain rule, ddxlnu=uu.\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 (x2)4(x+1)3(x3)6,\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=(x2)4(x+1)3(x3)6y=\dfrac{(x-2)^4(x+1)^3}{(x-3)^6} lny=ln(x2)4(x+1)3(x3)6\ln y=\ln \dfrac{(x-2)^4(x+1)^3}{(x-3)^6} lny=4ln(x2)+3ln(x+1)6ln(x3)\ln y=4\ln(x-2)+3\ln(x+1)-6\ln(x-3) 1ydydx=4x2+3x+16x3\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{4}{x-2}+\dfrac{3}{x+1}-\dfrac{6}{x-3} dydx=y(4x2+3x+16x3)\dfrac{dy}{dx}=y\left(\dfrac{4}{x-2}+\dfrac{3}{x+1}-\dfrac{6}{x-3}\right) dydx=((x2)4(x+1)3(x3)6)(4x2+3x+16x3)\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.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 (x+1)4(x2)3.\frac{(x+1)^4}{(x-2)^3}.


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

Let's start with the product rule. Let F=fgF=fg F=fgF=fg lnF=lnfg\ln F=\ln fg lnF=lnf+lng\ln F=\ln f+\ln g FF=ff+gg\dfrac{F'}{F}=\dfrac{f'}{f}+\dfrac{g'}{g} FF=fg+gffg\dfrac{F'}{F}=\dfrac{f'g+g'f}{fg} F=F(fg+gffg)F'=F\left(\dfrac{f'g+g'f}{fg}\right) F=fg+gfF'=f'g+g'f So this is the proof of the product rule. Now let's move on the quotient rule. Now let F=fg.F=\dfrac{f}{g}. F=fgF=\dfrac{f}{g} lnF=lnfg\ln F=\ln\dfrac{f}{g} lnF=lnflng\ln F=\ln f-\ln g FF=ffgg\dfrac{F'}{F}=\dfrac{f'}{f}-\dfrac{g'}{g} FF=fggffg\dfrac{F'}{F}=\dfrac{f'g-g'f}{fg} F=F(fggffg)F'=F\left(\dfrac{f'g-g'f}{fg}\right) F=(fg)(fggffg)F'=\left(\dfrac{f}{g}\right)\left(\dfrac{f'g-g'f}{fg}\right) F=fggfg2F'=\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.
5 years, 6 months ago

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Can you use similar techniques to prove that ddxxn=nxn1?\frac{d}{dx}x^n=nx^{n-1}?

Trevor B. - 5 years, 6 months ago

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Let y=xny = x^n. Taking the natural logarithm of both sides gives lny=nlnx\ln y = n \ln x . Differentiating, one has ddydydxlny=nddxlnx\frac{d}{dy} \cdot \frac{dy}{dx} \ln y = n \frac{d}{dx} \ln x which simplifies to dydx1y=nx\frac{dy}{dx}\frac{1}{y} = \frac{n}{x}. Substituting y=xny = x^{n} back into the equation and rearranging, we find that dydx=nxn1\frac{dy}{dx} = nx^{n-1}, as desired.

Lee Wall - 5 years, 6 months ago

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Good job!

Trevor B. - 5 years, 6 months ago

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@Trevor B. Thanks! It took me a bit too long to format it, though...

Lee Wall - 5 years, 6 months ago

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Yup, we can.

Let's say y=xny = x^n

ddxy=ddxxn\frac{d}{dx} y = \frac{d}{dx} x^n

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

ddxlny=ddxlnxn=ddxnlnx\frac{d}{dx} lny = \frac{d}{dx} ln x^n = \frac{d}{dx} nln x

1ydydx=n1x\frac{1}{y} \frac{dy}{dx} = n\frac{1}{x}

Multiply both sides by yy.

y1ydydx=ny1xy\frac{1}{y} \frac{dy}{dx} = ny\frac{1}{x}

And so we get...

dydx=ny1x\frac{dy}{dx} = ny\frac{1}{x} and remember, y=xny = x^n, hence substituting it we get,

dydx=nxn1x\frac{dy}{dx} = nx^n\frac{1}{x}

Using exponent properties, I can say...

dydx=nxnx1\frac{dy}{dx} = nx^n x^{-1}

dydx=nxn1\frac{dy}{dx} = nx^{n-1}

y=xnddxxn=nxn1y = 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. ^^

Vishnuram Leonardodavinci - 5 years, 6 months ago

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Good job!

Trevor B. - 5 years, 6 months ago

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Can anyone think of what else you can use logarithmic differentiation for (in terms of proofs)?

Trevor B. - 5 years, 6 months ago

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Something like xxx^x or even xxxx^{x^x}

Vishnuram Leonardodavinci - 5 years, 6 months ago

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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)

Abhinav Raichur - 5 years, 2 months ago

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@Trevor B. Can you add these to the Differentiation Rules Wiki pages? Thanks!

Calvin Lin Staff - 4 years, 11 months ago

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Luv u de

Swastik Behera - 3 years, 11 months ago

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