Table of Contents

Last modified on May 24th, 2024

Finding the derivative of any logarithmic function is called logarithmic differentiation.

The derivative of the natural logarithmic function (with the base ‘e’), lnx, with respect to ‘x,’ is ${\dfrac{1}{x}}$ and is given by

${\dfrac{d}{dx}\left( \ln x\right) =\left( \ln x\right)’=\dfrac{1}{x}}$, where x > 0

Similarly, the derivative of the logarithmic functions to the base ‘b,’ log_{b}x, with respect to ‘x,’ called the common logarithm ${\dfrac{1}{x\ln b}}$ is represented by

${\dfrac{d}{dx}\left( \log _{b}x\right) =\left( \log _{b}x\right)’=\dfrac{1}{x\ln b}}$, where x > 0 and ‘b’ is any positive real number except 1 (b Є ℝ^{+} and b ≠ 1)

To prove the derivative of the natural logarithmic function, we use the implicit differentiation of its inverse, also known as the exponential form.

Let us assume y = lnx = log_{e}x.

Converting it into its exponential form, we get

e^{y} = x

Now, by differentiating both sides with respect to x, we get

${\dfrac{d}{dx}\left( e^{y}\right) =\dfrac{d}{dx}\left( x\right)}$

⇒ ${\dfrac{d}{dx}\left( e^{y}\right) =1}$

⇒ ${e^{y}\ln e\dfrac{dy}{dx}=1}$

Since e^{y} = x, we have

${x\ln e\dfrac{dy}{dx}=1}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{x\ln e}}$

Since lne = 1, ${\dfrac{dy}{dx}=\dfrac{1}{x}}$

Mathematically, if h(x) = ln(f(x)), then for all values of x, f(x) > 0, and

${h’\left( x\right) =\dfrac{f’\left( x\right) }{f\left( x\right)}}$

The graph of y = log_{e}x = lnx and its derivative ${\dfrac{dy}{dx}=\dfrac{1}{x}}$ are shown below.

**Differentiate y = ln(2x sinx)**

Solution:

As we know, ${\dfrac{d}{dx}\left( \ln x\right) =\dfrac{1}{x}}$

Here, ${y=\ln \left( 2x\sin x\right)}$

⇒ ${\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \ln \left( 2x\sin x\right) \right]}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{2x\sin x}\left( \dfrac{d}{dx}\left( 2x\sin x\right) \right)}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{2x\sin x}\left( 2\sin x+2x\cos x\right)}$

⇒ ${\dfrac{dy}{dx}=\dfrac{2\left( \sin x+x\cos x\right) }{2x\sin x}}$

⇒ ${\dfrac{dy}{dx}=\dfrac{\left( \sin x+x\cos x\right) }{x\sin x}}$

To prove the common logarithmic function’s derivative, we use implicit differentiation similar to that used to prove the natural logarithmic function’s derivative.

Let us assume a logarithmic function y = log_{b}x

Converting it into its exponential form, we get

b^{y} = x

Now, by differentiating both sides with respect to x, we get

${\dfrac{d}{dx}\left( b^{y}\right) =\dfrac{d}{dx}\left( x\right)}$

⇒ ${\dfrac{d}{dx}\left( b^{y}\right) =1}$

⇒ ${b^{y}\ln b\dfrac{dy}{dx}=1}$

Since b^{y} = x, we have

${x\ln b\dfrac{dy}{dx}=1}$

⇒ ${\dfrac{dy}{dx}=\dfrac{1}{x\ln b}}$

Mathematically, if h(x) = log_{b}(f(x)), then for all values of x, f(x) > 0, and

${h’\left( x\right) =\dfrac{f’\left( x\right) }{f\left( x\right) \ln b}}$

We can also obtain the derivative formula of the common logarithmic function by using ${\dfrac{d}{dx}\left( \ln x\right) =\dfrac{1}{x}}$

Let us verify this by considering the function f(x) = log_{b}x.

Now, by changing the base rule, we get

${\log _{b}x=\dfrac{\log _{e}x}{\log _{e}b}}$

Since logₑ = ln, f(x) = ${\dfrac{\ln x}{\ln b}}$

Differentiating both sides with respect to x, we get

${f’\left( x\right) =\dfrac{d}{dx}\left( \dfrac{\ln x}{\ln b}\right)}$

⇒ ${f’\left( x\right) =\left( \dfrac{1}{\ln b}\right) \dfrac{d}{dx}\left( \ln x\right)}$

⇒ ${f’\left( x\right) =\left( \dfrac{1}{\ln b}\right) \left( \dfrac{1}{x}\right)}$

⇒ ${\dfrac{d}{dx}\left( \log _{b}x\right) =\dfrac{1}{x\ln b}}$

Thus, the derivative of log_{b}x with respect to x is ${\dfrac{1}{x\ln b}}$, proved.

The graph of y = log_{10}x = logx and its derivative ${\dfrac{dy}{dx}=\dfrac{1}{x\ln 10}}$ are shown below.

**Differentiate y = log _{4}(5x^{2} + 7)**

Solution:

As we know, ${\dfrac{d}{dx}\left( \log _{b}x\right) =\dfrac{1}{x\ln b}}$

Here, ${y=\log _{4}\left( 5x^{2}+7\right)}$

⇒ ${\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \log _{4}\left( 5x^{2}+7\right) \right]}$

⇒ ${\dfrac{dy}{dx}=\left( \dfrac{1}{5x^{2}+7}\right) \left( \dfrac{1}{\ln 4}\right) \dfrac{d}{dx}\left( 5x^{2}+7\right)}$

⇒ ${\dfrac{dy}{dx}=\left( \dfrac{1}{5x^{2}+7}\right) \left( \dfrac{1}{\ln 2^{2}}\right) \left( 10x\right)}$

⇒ ${\dfrac{dy}{dx}=\left( \dfrac{1}{5x^{2}+7}\right) \left( \dfrac{1}{2\ln 2}\right) \left( 10x\right)}$ (by the power rule of the logarithm)

⇒ ${\dfrac{dy}{dx}=\left( \dfrac{1}{5x^{2}+7}\right) \left( \dfrac{1}{\ln 2}\right) \left( 5x\right)}$

⇒ ${\dfrac{dy}{dx}=\dfrac{5x}{\left( \ln 2\right) \left( 5x^{2}+7\right) }}$

Last modified on May 24th, 2024