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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, 10 months ago

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Can you use similar techniques to prove that \(\frac{d}{dx}x^n=nx^{n-1}?\) Trevor B. · 2 years, 10 months ago

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@Trevor B. 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. Lee Wall · 2 years, 10 months ago

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@Lee Wall Good job! Trevor B. · 2 years, 10 months ago

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@Trevor B. Thanks! It took me a bit too long to format it, though... Lee Wall · 2 years, 10 months ago

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@Trevor B. 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. ^^ Vishnuram Leonardodavinci · 2 years, 10 months ago

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@Vishnuram Leonardodavinci Good job! Trevor B. · 2 years, 10 months ago

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Can anyone think of what else you can use logarithmic differentiation for (in terms of proofs)? Trevor B. · 2 years, 10 months ago

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@Trevor B. Something like \(x^x\) or even \(x^{x^x}\) Vishnuram Leonardodavinci · 2 years, 10 months ago

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@Vishnuram Leonardodavinci 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 · 2 years, 7 months ago

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Luv u de Swastik Behera · 1 year, 3 months ago

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@Trevor B. Can you add these to the Differentiation Rules Wiki pages? Thanks! Calvin Lin Staff · 2 years, 3 months ago

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