# Logarithmic Differentiation – Examples and Formula

Here you will learn formula of logarithmic differentiation with examples.

Let’s begin –

## Logarithmic Differentiation

We have learnt about the derivatives of the functions of the form $$[f(x)]^n$$ , $$n^{f(x))}$$ and $$n^n$$ , where f(x) is a function of x and n is a constant. In this section, we will be mainly discussing derivatives of the functions of the form $$[f(x)]^{g(x)}$$ where f(x) and g(x) are functions of x x. To find the derivative of this type of functions we proceed as follows :

Let y = $$[f(x)]^{g(x)}$$. Taking logarithm of both the sides, we get

log y = g(x) . log{f(x)}

Differrentiating with respect to x, we get

$$1\over y$$ $$dy\over dx$$ = g(x) $$\times$$ $$1\over f(x)$$ $$d\over dx$$ ((f(x)) + log {f(x)}.$$d\over dx$$(g(x))

$$\therefore$$  $$dy\over dx$$ = y{$${g(x)\over f(x)}$$.$$d\over dx$$(f(x)) + log{f(x)}.$$d\over dx$$ (g(x))}

Alternatively, we may write

y = $$[f(x)]^{g(x)}$$ = $$e^{g(x)log{f(x)}}$$

Differentiating with respect to x, we get

$$dy\over dx$$ = $$e^{g(x)log{f(x)}}$$ { g(x) $$\times$$ $$1\over f(x)$$ $$d\over dx$$ ((f(x)) + log {f(x)}.$$d\over dx$$(g(x)) }

$$\implies$$ $$dy\over dx$$ = $$[f(x)]^{g(x)}$${$${g(x)\over f(x)}$$.$$d\over dx$$(f(x)) + log{f(x)}.$$d\over dx$$ (g(x))}

Example : Differentiate $$x^x$$ with respect to x.

Solution : Let y = $$x^x$$. Then,

Taking log both sides,

log y = x.log x

$$\implies$$ y = $$e^{x.log x}$$

On differentiating both sides with respect to x, we get

$$dy\over dx$$ = $$e^{x.log x}$$$$d\over dx$$(xlogx)

$$\implies$$ $$dy\over dx$$ = $$x^x{log x \times {d\over dx}(x) + x \times {d\over dx}(log x)}$$

= $$x^x(log x + x\times {1\over x})$$

$$\implies$$ $$dy\over dx$$ = $$x^x(1 + logx)$$

Example : Differentiate $$x^{sinx}$$ with respect to x.

Solution : Let y = $$x^{sinx}$$. Then,

Taking log both sides,

log y = sin x.log x

$$\implies$$ y = $$e^{sin x.log x}$$

On differentiating both sides with respect to x, we get

$$dy\over dx$$ = $$e^{sin x.log x}$$$$d\over dx$$(sin x.log x)

$$\implies$$ $$dy\over dx$$ = $$x^{sin x}{log x {d\over dx}(sin x) + sin x {d\over dx}(log x)}$$

$$\implies$$ $$dy\over dx$$ = $$x^{sin x}(cos x.log x + {sin x\over x}$$)