DIFFERENTIATION OF LOGARITHMIC FUNCTION

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We can differentiate logarithmic functions using the property of logs.

Recalled that $\log_ax$ can be rewritten as:

$$\log_ax=\frac{\log_ex}{\log_ea}$$

If $log_e=\ln$, then

$$\log_ax=\frac{in(x)}{\ln(a)}$$

Therefore, 

$$\frac{d}{dx}(\log_ax)=\frac{d}{dx}\left(\frac{\ln(x)}{\ln(a)}\right)$$.

Using this fact, We can obtain a general expression for differentiating logarithmic function as exemplified as 

$$\frac{d}{dx}(\log_{f(x)}{g(x)})=\frac{d}{dx}\left(\frac{\ln(g(x)}{\ln(f (x)}\right)$$

Example 1

Differentiate $y=\log_2(1-3x)$

Solution:

Recall that $\frac{d}{dx}(\log_ax)=\frac{d}{dx}\left(\frac{\ln(x)}{\ln(a)}\right)$

Therefore,

$$\frac{d}{dx}(\log_{2}(1-3x)=\frac{d}{dx}\left(\frac{\ln(1-3x)}{\ln(2)}\right)$$

But $\frac{\ln(1-3x)}{\ln(2)}$ is a quotient, therefore, we use the quotient rule which is exemplified as:

$$\left(\frac{f}{g}\right)'=\frac{f'g-g'f}{(g)^2}$$

Here, $f=\ln(1-3x)$, $f'=\frac{-3}{1-3x}$, $g=\ln(2)$, $g'=0$

$$\frac{dy}{dx}=\frac{\frac{-3}{1-3x}\times \ln(2)-0(\ln(1-3x)}{(\ln(2))^2}$$

$$\frac{dy}{dx}=\frac{\frac{3\ln(2)}{1-3x}}{(\ln(2))^2}$$

$$\frac{dy}{dx}=\frac{3 \ln(2)}{1-3x}\div (\ln(2))^2$$

$$\frac{dy}{dx}=\frac{3 \ln(2)}{1-3x}\times\frac{1}{(\ln(2))^2}$$

$$\frac{dy}{dx}=\frac{-3}{(1-3x)(\ln(2)}$$

$$\frac{dy}{dx}=\frac{-3}{\ln (2)-3x( \ln(2))}$$

Example 2

Differentiate $y=\log_{10}\sqrt{x}$

Solution:

Recall that $\frac{d}{dx}(\log_ax)=\frac{d}{dx}\left(\frac{\ln(x)}{\ln(a)}\right)$

Therefore, 

$$\frac{dy}{dx}(\log_{10}\sqrt{x})=\frac{dy}{dx}\left(\frac{\ln(\sqrt{2})}{\ln(10)}\right)$$

Remember quotient rule:

$$\left(\frac{f}{g}\right)'=\frac{f'g-g'f}{(g)^2}$$

$f=\ln\sqrt{2}$, $f'=\frac{1}{2x}$, $g=\ln(10)$, $g'=0$

$$\frac{dy}{dx}=\frac{\frac{1}{2x}(\ln(10))-0(\ln(\sqrt{x})}{(\ln(10))^2}$$

$$\frac{dy}{dx}=\frac{\frac{\ln(10)}{2x}}{(\ln(10))^2}$$

$$\frac{dy}{dx}=\frac{\ln(10)}{2x}\times\frac{1}{(\ln(10))^2}$$

$$\frac{dy}{dx}=\frac{1}{2x(\ln(10))}$$

Example 3

Differentiate $y=\log_2x^2-\log24x$

Solution:

From the logarithm rules,

$$\log_xa-\log_xb=\log_x\left(\frac{a}{b}\right)$$

Therefore, $\log_2x^2-\log_24x=\log_2\left(\frac{x^2}{4x}\right)=\log_2\left(\frac{x}{4}\right)$

$$\frac{d}{dx}(\log_2\left(\frac{x}{4}\right)=\frac{d}{dx}(\left(\frac{\ln(\frac{x}{4})}{\ln(2)}\right)$$

Here, $f=\ln(\frac{x}{4})$, $f'=\frac{1}{x}$, $g=\ln(2)$, $g'=0$.

$$\frac{dy}{dx}=\frac{\frac{1}{x}(\ln(2))-0(\ln(\frac{x}{4})}{(\ln(2))^2}$$

$$\frac{dy}{dx}=\frac{\ln(2)}{x}÷(\ln(2))^2$$

$$\frac{dy}{dx}=\frac{\ln(2)}{x}×\frac{1}{(\ln(2))^2}$$

$$\frac{dy}{dx}=\frac{1}{x (\ln(2))}$$

Example 4

Differentiate $\ln(e^2)$

Solution

If $\ln=\log_e$, Then

$in(e^2)=\log_ee^2=2(1)=2$

$\frac{d}{dx}(2)=0$

Remember that the derivative of a constant is zero.

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Example 5

If $y=\log_4(8x^2-6x-10)$

Solution

$\frac{d}{dx}(\log_4(8x^2-6x-10))=\frac{\ln(8x^2-6x-10)}{\ln(4)}$

Now, Let differentiate $\frac{\ln(8x^2-6x-10)}{\ln(4)}$ via quotient rule.

But First, we need to obtain the derivative of $\ln(8x^2-6x-10)$

Previously, I told you that the general expression for differentiation of natural logs is:

$(\ln(f(x))'=\frac{f'(x)}{f(x)}$

$$(\ln(8x^2-6x-10))'=\frac{16x-6}{8x^2-6x-10}$$

Using quotient rule, $f=\ln(8x^2-6x-10)$, $f'=\frac{16x-6}{8x^2-6x-10}$, $g=\ln(4)$, $g'=0$.

$$\frac{dy}{dx}=\frac{(\frac{16x-6}{8x^2-6x-10})(\ln(4)-0(\ln(8x^2-6x-10)}{(\ln(4))^2}$$

$$\frac{dy}{dx}=\frac{\frac{(16x-6)(\ln(4))}{8x^2-6x-10}}{(\ln(4))^2}$$

$$\frac{dy}{dx}=\frac{(16x-6)(\ln(4))}{8x^2-6x-10}÷(\ln(4))^2$$

$$\frac{dy}{dx}=\frac{(16x-6)(\ln(4))}{8x^2-6x-10}×\frac{1}{(\ln(4))^2}$$

$$\frac{dy}{dx}=\frac{16x-6}{(8x^2-6x-10)(\ln(4))}$$

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