# 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*

## General Result

Prove that:\[\dfrac{d}{dx} \log_{a} x = \dfrac{1}{x\ln a}\]

Proof:Let \(y=\log_{a} x\) and 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:

\[a^y\ln a\dfrac{dy}{dx}=1\]

\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{a^y\ln a}=\dfrac{1}{x\ln a}\]

## Special case of the General Result

*Image Source: Wikipedia*

If we put \(a=e\) in the General Result where \(e\) is the Euler's number , we have \(\ln a=\ln e=1\) and we 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:\[\dfrac{d}{dx} \ln |x|=\dfrac{1}{x}\]

Proof:Using the property of Absolute Value , we know this 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,

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

\[\Rightarrow f'(x)=\dfrac{1}{x} \]

\[\Rightarrow \dfrac{d}{dx} \ln|x|=\dfrac{1}{x} \ \forall \ x \neq 0\]

## Example Problems

Prove the Power rule using Logarithmic Differentiation.

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 \(\dfrac{d}{dx} \ln|x|=\dfrac{1}{x}\) (as proved above) we have:

\[\dfrac{y'}{y}=\dfrac{n}{x}\]

\[\Rightarrow y'=\dfrac{ny}{x}=\dfrac{nx^{n}}{x}=nx^{n-1}\]

which was to be proved.

Differentiate: \(y=x^{\sqrt{x}}\)

Solution: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:

\[\dfrac{d}{dx} \sqrt{x}\ln x = \sqrt{x} \cdot \dfrac{1}{x} + \ln x \cdot \dfrac{1}{2\sqrt{x}}=\dfrac{2+\ln x}{2\sqrt{x}} \]

Thus we have:

\[\dfrac{d}{dx} x^{\sqrt{x}}= x^{\sqrt{x}}\left(\dfrac{2+\ln x}{2\sqrt{x}}\right)\]

Differentiate: \(y=2e^x+5x^2\ln x\)

Solution: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.

Differentiate: \(y=\dfrac{2e^x}{3e^x+1}\)

Solution:This is where we need to directly use quotient rule.Using Quotient Rule we have,

\[\dfrac{dy}{dx}=\dfrac{2e^x(3e^x+1)-(3e^x)(2e^x)}{(3e^x+1)^2}= \dfrac{6e^{2x}+2e^x-6e^{2x}}{(3e^x+1)^2}\]

\[\Rightarrow \dfrac{dy}{dx}=\dfrac{2e^x}{(3e^x+1)^2}\]

Differentiate: \(y=x^x\) for \(x>0\)

Solution:We cannot directly approach this using differentiation rules.We need to bring suitable form for the function to be differentiated:\[y=x^x\Rightarrow \ln y=\ln x^x \Rightarrow \ln y= x\ln x\]

We now differentiate both sides with respect to \(x\), using Chain rule on the left side and the Product rule on the right:

\[\dfrac{dy}{dx} \cdot \dfrac{1}{y} = \ln x + x \cdot\dfrac{1}{x}=\ln x + 1\]

\[\Rightarrow \dfrac{dy}{dx}= y(\ln x + 1) = x^x(\ln x+1)\]

## Community problems

**Cite as:**Applying Differentiation Rules To Logarithmic Functions.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/applying-differentiation-rules-to-logarithmic/