DIFFERENTIATION OF NATURAL LOGARITHMS (ln)

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The natural logarithm of a number is its logarithm to the base of e.

The natural logarithm of x  is commonly written as $\log_{e}x$, $\ln (x)$.

To differentiate natural logs, The general expression is:

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

For example, to differentiate $\ln (x)$, it will be the derivative of x (which is 1), divided by x. That is,

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

Example 1

Differentiate $4 \ln x+2$

Solution:

As I told you before, $\frac{d}{dx}(\ln (f(x))=\frac{f'(x)}{f(x)}$

Therefore

$$\frac{d}{dx}(4 \ln (x) +2)=4(\frac{1}{x})$$

Note, the derivative of a constant (2) is zero.

$$\frac{d}{dx}(4 \ln (x)+2)=\frac{4}{x}$$

Example 2

Differentiate $y=\ln (x^2+4)$

Solution:

The derivative of x²+4 is 2x. Hence

$$\frac{dy}{dx}=\frac{2x}{x^2+4}$$

Example 3

Differentiate $y=\ln(\frac{1}{x^2+9})$

Solution:

Recalled that:

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

Here, $f(x)=\frac{1}{x^2+9}$, $f'(x)=\frac{-2x}{(x^2+9)^2}$, 

$$\frac{dy}{dx}=\frac{\frac{-2x}{(x^2+9)^2}}{\frac{1}{x^2+9}}$$

$$\frac{dy}{dx}=\frac{-2x}{(x^2+9)^2}\div\frac{1}{x^2+9}$$

$$\frac{dy}{dx}=\frac{-2x}{(x^2+9)^2}\times\frac{x^2+9}{1}$$

$$\frac{dy}{dx}=\frac{-2x}{x^2+9}$$

Example 4

If $h(x)=(x+1)(\ln (x))$, Find $h'(x)$

Solution:

This involves a product, therefore, we use the product rule. Recalled product rule:

$$(fg)'=f'g+g'f$$

f=x+1, f'=1, $g=\ln (x)$, $g'=\frac{1}{x}$

$$h'(x)=1(\ln (x))+\frac{1}{x}(x+1)$$

$$h'(x)=\frac{\ln (x)}{1}+\frac{x+1}{x}$$

$$h'(x)=\frac{x \ln (x)+x+1}{x}$$

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Example 5

Differentiate $y=\frac{\ln (x)}{1}-\frac{x}{4}$

Solution:

$$y=\frac{\ln (x)}{1}-\frac{x}{4}$$

Taking the L.C.M

$$y=\frac{4 \ln (x)-x}{4}$$

Now, we have a quotient, therefore, we use the quotient rule of differentiation, which is exemplified as:

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

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

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

$$\frac{dy}{dx}=\frac{\frac{16}{x}-\frac{4}{1}}{16}$$

$$\frac{dy}{dx}=\frac{16-4x}{x}÷16$$

$$\frac{dy}{dx}=\frac{16-4x}{x}\times\frac{1}{16}$$

$$\frac{dy}{dx}=\frac{4(4-x)}{x}\times\frac{1}{16}$$

$$\frac{dy}{dx}=\frac{4-x}{x}\times\frac{1}{4}=\frac{4-x}{4x}$$

Voila, We just learned the differentiation of natural log. Next, we will be looking at the differentiation of logarithmic functions.

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