There are cases in which differentiating the logarithm of a given function is simpler as compared to differentiating the function itself. By the proper usage of properties of logarithms and chain rule finding, the derivatives becomes easy.This concept is applicable to nearly all the non-zero functions which are differentiable in nature.

Therefore, in calculus the differentiation of some complex functions is done by taking logarithms and then the logarithmic derivative is utilized to solve such a function.

The equations which take the form y = f(x) = [u(x)]^{{v(x)}} can be easily solved using the concept of logarithmic differentiation.

For differentiating functions of this type we take on both the sides of the given equation.

Therefore, taking log on both sides we get,log y = log[u(x)]^{{v(x)}}

log y = v(x)log u(x)

Now, differentiating both the sides w.r.t. x by implementing chain rule, we get

The only constraint for using logarithmic differentiation rules is that f(x) and u(x) must be positive as logarithmic functions are only defined for positive values.

The basic properties of real logarithms are generally applicable to the logarithmic derivatives.

For example: (log uv)’ = (log u + log v)’ = (log u)’ + (log v)’

Now, as we are thorough with logarithmic differentiation rules let us take some logarithmic differentiation examples to know a little bit more about this.

__Example 1:__Find the value of \(\frac{dy}{dx}\) if,\(y = e^{x^{4}}\)

__Solution 1:__ Given the function \(y = e^{x^{4}}\)

Taking natural logarithm of both the sides we get,

ln y = ln e^{x4}

ln y = x^{4} ln e

ln y = x^{4}

Now, differentiating both the sides w.r.t we get,

\(\frac{1}{y} \frac{dy}{dx}\) = \(4x^3 \)

\( \Rightarrow \frac{dy}{dx}\) =\( y.4x^3\)

\(\Rightarrow \frac{dy}{dx}\) =\( e^{x^{4}}×4x^3\)

Therefore, we see how easy and simple it becomes to differentiate a function using logarithmic differentiation rules.

__Example 2:__Find the value of \(\frac{dy}{dx}\) if y = 2x^{{cos x}}.

__Solution 2:__Given the function y = 2x^{{cos x}}

Taking logarithm of both the sides, we get

log y = log(2x^{{cos x}})

\(\Rightarrow log y = log 2 + log x^{cos x} (As log(mn) = log m + log n)\)

\(\Rightarrow log y = log 2 + cos x × log x (As log m^n = log m + log n)\)

Now, differentiating both the sides w.r.t by using the chain rule we get,

\(\frac{1}{y} \frac{dy}{dx} = \frac{cos x}{x} – (sin x)(log x)\)

This is yet another equation which becomes simplified after using logarithmic differentiation rules.

We have seen how useful it can be to use logarithms to simplify differentiation of various complex functions.Welcome to the world of BYJU’s to get to know more about differential calculus, download Byjus the learning app.