 # Logarithmic Differentiation

Logarithmic differentiation is a very useful method to differentiate some complicated functions which can’t be easily differentiated using the common techniques like the Chain Rule. It can also be employed for the functions that involve many terms that need the application of the Product Rule or the Quotient Rule multiple times to be differentiated.

This technique greatly simplifies the process of differentiation as well as the solution so obtained. In physics, error calculations for experiments are usually done by the log-error method, not the normal derivative method. Let’s find out more about the logarithmic differentiation in the section below.

### Suggested Videos        Continuity of Functions Introduction to Derivatives Derivatives of Trigonometric Functions using First Principle ## The Method of Logarithmic Differentiation

Follow the following steps to find the differentiation of a logarithmic function:

• Take the natural logarithm of the function to be differentiated.
• Use the properties of logarithmic functions to distribute the terms that were initially accumulated together in the original function and were tough to differentiate.
• Differentiate the resulting equation.
• Multiply the equation by the function itself to get the derivative.

Now let us understand the working of the method by solving a couple of problems!

## Solved Examples for you

Question: Compute the derivative of the function $${ y = \frac{(x^2 + 1).(x – 5)}{(x^3 +2)(x + 5)} }$$.

Solution: Note that if you start using the Product Rule and the Quotient Rule of Differentiation here, it would be a very lengthy process of obtaining the derivative. By the method of logarithmic differentiation, we’ll save a lot of time. Let’s see how –

Taking the natural logarithm –
$${ ln(y) = ln (\frac{(x^2 + 1).(x – 5)}{(x^3 +2)(x + 5)}) }$$
Using the properties of logarithms –
$${ ln(y) = ln (x^2 + 1) + ln(x – 5) – ln(x^3 +2) – ln(x + 5) }$$
Now, on differentiating the equation with respect to x, we’ll get –
$${ \frac{d}{dx}(ln(y)) = \frac{d}{dx}(ln (x^2 + 1) + ln(x – 5) – ln(x^3 +2) – ln(x + 5)) }$$
$${ : \frac{1}{y}.\frac{dy}{dx} = \frac{1}{x^2 + 1}.(2x) + \frac{1}{x – 5}.(1) – \frac{1}{x^3 +2}.(3x^2) – \frac{1}{x + 5}.(1) }$$
Multiplying by y on both sides, and substituting the value of y, we get –
$${ \frac{dy}{dx} = \frac{(2x).(x – 5)}{(x^3 +2)(x + 5)} + \frac{x^2 + 1}{(x^3 +2)(x + 5)} – \frac{(x^2 + 1).(x – 5).(3x^2)}{(x^3 +2)^2(x + 5)} – \frac{(x^2 + 1).(x – 5)}{(x^3 +2)(x + 5)^2} }$$
which is the required result.

Question: Compute the derivative of the function $${y = x^{cos^2x}}$$.

Solution: Note that there is no known formula which can be directly employed to differentiate the above function. See how the technique of logarithmic differentiation saves us here.

Taking the natural logarithm –
$${ln(y) = ln(x^{cos^2x})}$$
Using the properties of the logarithms –
$${ln(y) = (cos^2x).ln(x)}$$
Differentiate with respect to x, and use the chain rule on the right hand side –
$${\frac{d}{dx}(ln(y)) = (cos^2x).\frac{d}{dx}(ln(x)) + \frac{d}{dx}(cos^2x).ln(x)}$$
$${ : \frac{1}{y}.\frac{dy}{dx} = (cos^2x).\frac{1}{x} + 2cosx.(-sinx).ln(x)}$$
Multiplying by y on both sides, and substituting the value of y, we get –
$${ \frac{dy}{dx} = x^{cos^2x – 1}.(cos^2x) – 2sinx.cosx.ln(x).x^{cos^2x}}$$
$${ \frac{dy}{dx} = x^{-sin^2x}.(cos^2x) – sin2x.ln(x).x^{cos^2x}}$$

This should give you a pretty good idea about how to apply this method of differentiation to any problem you encounter on the logarithmic functions.

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