Derivatives in maths enable the calculation of function change rates in terms of variables. The function defined by \(y=\log_{a}(x)\) (\(x>0\)) where \(x=a^{y}\), \(a>0\), \(a\neq 1\) is called the logarithm of x to the base a. The derivative of \(\log_{a}(x)\) is \(\frac{1}{x\ln(a)}\). The derivative of \(\log_{a}(x)\) is denoted by \(\frac{\mathrm{d}}{\mathrm{d}x}(\log_{a}(x))\) or \((\log_{a}(x))’\).

We will learn the concept of derivative of \(\log(x)\), proof of derivative of \(\log(x)\) by first principle, by implicit differentiation, by using derivative of \(\ln(x)\) and solve problems based on derivative of \(\log(x)\).

What is Derivative of Log x?

Derivative of \(\log_{a}(x)\) is \(\frac{1}{x\ln(a)}\). Here “\(\ln\)” is the derivative of “\(\log\)”. “\(\ln\)” is called the natural logarithm or it is a logarithm with base ‘\(e\)’, i.e. \(\ln=\log_{e}\).

In general, a logarithm has the form \(\log_{a}(x)\). That is, we call \(a\) the base of the logarithm. Also, \(\log_{a}(x)\) represents the number we raise \(a\) to in order to get \(x\). Now, notice that \(\log(x)\) doesn’t have a base shown. When this is the case, the implied base is \(10\). Therefore, \(\log(x) = \log_{10}(x)\). So the derivative of \(\log(x)\) is \(\frac{1}{x\ln(10)}\).

The derivative of \(\log(x)\) is denoted by \(\frac{\mathrm{d}}{\mathrm{d}x}(\log(x))\) or \((\log(x))’\). Thus, we have

• \(\frac{\mathrm{d}}{\mathrm{d}x}(\log_{a}(x))\) or \((\log_{a}(x))’\) = \(\frac{1}{x\ln(a)}\).
• \(\frac{\mathrm{d}}{\mathrm{d}x}(\log(x))\) or \((\log(x))’\) = \(\frac{1}{x\ln(10)}\).

Derivative of Log x Proof by First Principle

To Prove: \(\frac{\mathrm{d}}{\mathrm{d}x}(\log_{a}(x))=\frac{1}{x\ln(a)}\) by using first principle.

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

By first principle, the derivative of a logarithmic function \(f(x)\) (which is denoted by \(f’(x)\) is given by the limit,

\(f'(x)=\displaystyle \lim_{h\to 0}\frac{[f(x+h)-f(x)]}{h}\)

Since \(f(x)=\log_{a}(x)\), we have \(f(x+h)=\log_{a}(x+h)\).

Substituting these values in the equation of first principle,

\(f'(x)=\displaystyle \lim_{h\to 0}\frac{[\log_{a}(x+h)-\log_{a}(x)]}{h}\)

By using the property of logarithms, \(\log_{a}(m)-\log_{a}(n)=\log_{a}\left(\frac{m}{n}\right)\), we get

\(f'(x)=\displaystyle \lim_{h\to 0}\frac{(\log_{a}[\frac{x+h}{x}])}{h}\)

\(\Rightarrow\) \(f'(x)=\displaystyle \lim_{h\to 0}\frac{(\log_{a}(1+(\frac{h}{x})))}{h}\)

Assume \(\frac{h}{x}=t\) \(\Rightarrow\) \(h=xt\), we get

When \(h\to 0\), \(\frac{h}{x}\to 0\) \(\Rightarrow\) \(t\to 0\)

Then the above limit becomes

\(f'(x)=\displaystyle \lim_{t\to 0}\frac{[\log_{a}(1+t)]}{xt}\)

\(\Rightarrow\) \(f'(x)= \displaystyle \lim_{t\to 0}\frac{1}{(xt)}\log_{a}(1+t)\)

By using property of logarithm, \(m\log_{a}(a)=\log_{a}(a^{m})\), we get

\(f'(x)=\displaystyle \lim_{t\to 0}\log_{a}(1+t)^{\frac{1}{(xt)}}\)

By using a property of exponents, \(a^{mn}=(a^{m})^{n}\), we get

\(f'(x)=\displaystyle \lim_{t\to 0}\log_{a}[(1+t)^{\frac{1}{t}}]^{\frac{1}{x}}\)

\(\Rightarrow\) \(f'(x)=\displaystyle \lim_{t\to 0}\left(\frac{1}{x}\right)\log_{a}[(1+t)^{\frac{1}{t}}]\)

Here, the variable of the limit is ‘\(t\)’. So we can write \(\left(\frac{1}{x}\right)\) outside of the limit.

\(f'(x)=\left(\frac{1}{x}\right)\displaystyle \lim_{t\to 0}\log_{a}[(1+t)^{\frac{1}{t}}]\)

\(\Rightarrow\) \(f'(x)=\left(\frac{1}{x}\right)\log_{a}\displaystyle \lim_{t\to 0}[(1+t)^{\frac{1}{t}}]\)

Using one of the formulas of limits, \(\displaystyle \lim_{t\to 0}[(1+t)^{\frac{1}{t}}]=e\).

\(\Rightarrow\) \(f'(x)=\left(\frac{1}{x}\right)\log_{a}(e)\)

\(\Rightarrow\) \(f'(x)=\left(\frac{1}{x}\right)\left(\frac{1}{\log_{e}(a)}\right)\) [\(\because\) ‘\(a\)’ and ‘\(e\)’ are interchanged]

\(\Rightarrow\) \(f'(x)=\left(\frac{1}{x}\right)\left(\frac{1}{\ln(a)}\right)\) [\(\because\) \(\log_{e}=\ln\)

\(\Rightarrow\) \(f'(x)=\frac{1}{(x\ln(a))}\)

Therefore, the derivative of \(\frac{\mathrm{d}}{\mathrm{d}x}(\log_{a}(x))\) is \(\frac{1}{x\ln(a)}\) by using first principle.

Derivative of Log x Proof by Implicit Differentiation

To Prove: \(\frac{\mathrm{d}}{\mathrm{d}x}(\log_{a}(x))=\frac{1}{x\ln(a)}\) by using implicit differentiation.

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

\(\Rightarrow\) \(a^{y}=x\)

Taking the derivative on both sides with respect to \(x\), we get

\(\frac{\mathrm{d}}{\mathrm{d}x}(a^{y})=\frac{\mathrm{d}}{\mathrm{d}x}(x)\)

By using the chain rule, we get

\(a^{y}\ln(a)\frac{\mathrm{d}y}{\mathrm{d}x}=1\)

\(\Rightarrow\) \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{a^{y}\ln(a)}\)

\(\because\) \(a^{y}=x\)

\(\Rightarrow\) \(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{x\ln(a)}\)

Therefore, the derivative of \(\frac{\mathrm{d}}{\mathrm{d}x}(\log_{a}(x))\) is \(\frac{1}{x\ln(a)}\) by using implicit function.

Derivative of Log x Proof using Derivative of ln x

To Prove: \(\frac{\mathrm{d}}{\mathrm{d}x}(\log_{a}(x))=\frac{1}{x\ln(a)}\) by using the derivative of \(\ln(x)\).

Note that the derivative of \(\ln(x)\) is \(\frac{1}{x}\). We can convert ‘\(\log\)’ into ‘\(\ln\)’ using change of base rule.

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

By using change of base rule, we have

\( f(x)=\frac{\log_{e}(x)}{\log_{e}(a)}\)

We know that \(\log_{e}=\ln\). Thus, we get

\(f(x)=\frac{\ln(x)}{\ln(a)}\)

Now find the derivative of \(f(x)\), then we get

\(f'(x)=\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{\ln(x)}{\ln(a)}\right]\)

\(\Rightarrow\) \(f'(x)=\frac{1}{\ln(a)}\frac{\mathrm{d}}{\mathrm{d}x}(\ln(x))\)

We know that \(\frac{\mathrm{d}}{\mathrm{d}x}(\ln(x)=\frac{1}{x}\), we get

\(\Rightarrow\) \(f'(x)=\frac{1}{\ln(a)}\cdot \frac{1}{x}\)

\(\Rightarrow\) \(f'(x)=\frac{1}{x\ln(a)}\)

Therefore, the derivative of \(\frac{\mathrm{d}}{\mathrm{d}x}(\log_{a}(x))\) is \(\frac{1}{x\ln(a)}\) by using derivative of \(\ln(x)\).

Solved Examples of Derivative of Log x

Example 1: Find the derivative of \(\log(x^{2})\).

Solution: Let \(y=\log(x^{2})\)

Using the derivatives of \(\log(x)\) formula, we have

\(\frac{\mathrm{d}}{\mathrm{d}x}(\log(x))=\frac{1}{x\ln(10)}\)

By using chain rule, we get

\(y’=\frac{1}{x^{2}\ln(10)}\frac{\mathrm{d}}{\mathrm{d}x}(x^{2})\)

\(\Rightarrow\) \(y’=\frac{1}{x^{2}\ln(10)}\cdot (2x)\)

\(\Rightarrow\) \(y’=\frac{2}{x\ln(10)}\)

Hence, the derivative of \(\log(x^{2})\) is \(\frac{2}{x\ln(10)}\).

Example 2: Find the derivative of \(\frac{x}{\log(x)}\).

Solution: Let \(y=\frac{x}{\log(x)}\)

Using the derivatives of \(\log(x)\) formula, we have

\(\frac{\mathrm{d}}{\mathrm{d}x}(\log(x))=\frac{1}{x\ln(10)}\)

By quotient rule, we have

\(\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{u}{v}\right]=\frac{v u’-u v’}{v^{2}}\)

\(\Rightarrow\) \(y’=\frac{\log(x)(1)-x(\frac{1}{x\ln(10)})}{(\log(x))^{2}}\)

\(\Rightarrow\) \(y’=\frac{\log(x)-(\frac{1}{\log(10)})}{(\log(x))^{2}}\)

Hence, the derivative of \(\frac{x}{\log(x)}\) is \(\frac{\log(x)-(\frac{1}{\log(10)})}{(\log(x))^{2}}\).

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