Applying Differentiation Rules To Logarithmic Functions

In this wiki, we will learn about differentiating logarithmic functions which are given by $y=\log_{a} x$ , in particular the natural logarithmic function $y=\ln x$ using the differentiation rules . We can easily prove that these logarithmic functions are easily differentiable by looking at there graphs:

Image Source: Wikipedia

Image Source: ltcconline.net

Prove that

$$\dfrac{d}{dx} \log_{a} x = \dfrac{1}{x\ln a}.$$

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Proof: Let $y=\log_{a} x,$ then by basic logarithmic properties we have

$$a^y=x.$$

Differentiating the above equation implicitly with respect to $x$ and using $\dfrac{d}{dx}a^x=a^x\ln a,$ we get

$$\begin{aligned} a^y\ln a\dfrac{dy}{dx}&=1\\ \Rightarrow \dfrac{dy}{dx}&=\dfrac{1}{a^y\ln a}\\&=\dfrac{1}{x\ln a}.\ _\square \end{aligned}$$

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If we put $a=e$ in the general result where $e$ is Euler's number, we have $\ln a=\ln e=1$ and get an amazing result:

$$\dfrac{d}{dx} \ln x=\dfrac{1}{x}.$$

But we have a more worth-remembering result than the above one:

Prove that

$$\dfrac{d}{dx} \ln |x|=\dfrac{1}{x}.$$

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Proof: Using the property of absolute value , we know the following about the function $f(x)=\ln|x|:$

$$f(x)=\begin{cases} \ln x \ & \text{if} \ & x>0 \\ \ln(-x) \ & \text{if} \ & x<0. \end{cases}$$

Now, it follows that

$$f'(x)=\begin{cases} \dfrac{1}{x} \ & \text{if} \ & x>0 \\ \dfrac{1}{-x} (-1)=\dfrac{1}{x} \ & \text{if} \ & x<0. \end{cases}$$

Therefore,

$$f'(x)=\dfrac{1}{x} \implies \dfrac{d}{dx} \ln|x|=\dfrac{1}{x} \ \forall \ x \neq 0.\ _\square$$

Prove the power rule using logarithmic differentiation.

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Proof:

First we state the power rule: for any real number $n$,

$$\frac{d}{dx} x^n = n x ^{n-1 }.$$

So first, we let $y=x^n,$ then by logarithmic differentiation for $x\neq 0$ we have

$$\ln|y|=\ln|x^n|=n\ln |x|.$$

Using $\frac{d}{dx} \ln|x|=\frac{1}{x}$ (as proved above), we have

$$\begin{aligned} \dfrac{y'}{y}&=\dfrac{n}{x}\\ \Rightarrow y'&=\dfrac{ny}{x}\\&=\dfrac{nx^{n}}{x}\\&=nx^{n-1}, \end{aligned}$$

which was to be proved. $_\square$

Differentiate $y=x^{\sqrt{x}}.$

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We can write $x=e^{\ln x}$ using logarithmic rules, and thus we have

$$x^{\sqrt{x}}=\left(e^{\ln x}\right)^{\sqrt{x}}.$$

Differentiating both the sides, we have

$$\dfrac{d}{dx} x^{\sqrt{x}}=\dfrac{d}{dx} e^{\sqrt{x}\ln x}.$$

Using chain rule, we have

$$\dfrac{d}{dx} e^{\sqrt{x}\ln x}=e^{\sqrt{x}\ln x}\dfrac{d}{dx} \sqrt{x}\ln x = x^{\sqrt{x}}\dfrac{d}{dx} \sqrt{x}\ln x.$$

Using product rule, we have

$$\dfrac{d}{dx} \sqrt{x}\ln x = \sqrt{x} \cdot \dfrac{1}{x} + \ln x \cdot \dfrac{1}{2\sqrt{x}}=\dfrac{2+\ln x}{2\sqrt{x}}.$$

Therefore, we have

$$\dfrac{d}{dx} x^{\sqrt{x}}= x^{\sqrt{x}}\left(\dfrac{2+\ln x}{2\sqrt{x}}\right).\ _\square$$

Differentiate $y=2e^x+5x^2\ln x.$

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First, let us try to differentiate the second term $5x^2\ln x.$

Using product rule,

$$\dfrac{d}{dx} 5x^2\ln x = 10x\ln x+5x^2\cdot\dfrac{1}{x}=10x\ln x+5x.$$

Thus, we have

$$\dfrac{dy}{dx}=2e^x+10x\ln x+5x,$$

which is what we needed to find. $_\square$

Differentiate $y=\dfrac{2e^x}{3e^x+1}.$

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This is where we need to directly use the quotient rule.

Using quotient rule, we have

$$\begin{aligned} \dfrac{dy}{dx} &=\dfrac{2e^x(3e^x+1)-(3e^x)(2e^x)}{(3e^x+1)^2}\\ &= \dfrac{6e^{2x}+2e^x-6e^{2x}}{(3e^x+1)^2}\\ \Rightarrow \dfrac{dy}{dx}&=\dfrac{2e^x}{(3e^x+1)^2}.\ _\square \end{aligned}$$

Differentiate $y=x^x$ for $x>0.$

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We cannot directly approach this using differentiation rules. We need to bring suitable form for the function to be differentiated:

$$y=x^x\implies \ln y=\ln x^x \implies \ln y= x\ln x.$$

We now differentiate both sides with respect to $x,$ using the chain rule on the left side and the product rule on the right:

$$\begin{aligned} \dfrac{dy}{dx} \cdot \dfrac{1}{y} &= \ln x + x \cdot\dfrac{1}{x}\\ &=\ln x + 1\\ \Rightarrow \dfrac{dy}{dx} &= y(\ln x + 1) \\ &= x^x(\ln x+1).\ _\square \end{aligned}$$