# Differentiation Using Logarithm

## Differentiation Using Logarithm

If y = [f₁(x)]f₂(x) (or) y = f₁(x) f₂(x) f₃(x) … (or) $$y=\frac{{{f}_{1}}(x)\ {{f}_{2}}(x)\ {{f}_{3}}(x)……}{{{g}_{1}}(x)\ {{g}_{2}}(x)\ {{g}_{3}}(x)……}$$.

(i) y = [f₁(x)]f₂(x)

log y = log [f₁(x)]f₂(x)

log y = f₂ (x) log [f₁ (x)]

(ii) y = f₁(x) f₂(x) f₃(x) …

log y = log [f₁(x) f₂(x) f₃(x) …]

log y = log f₁(x) + log f₂(x) + log f₃(x) + …

(iii) $$y=\frac{{{f}_{1}}(x)\ {{f}_{2}}(x)\ {{f}_{3}}(x)……}{{{g}_{1}}(x)\ {{g}_{2}}(x)\ {{g}_{3}}(x)……}$$,

$$\log y=\log \left( \frac{{{f}_{1}}(x)\ {{f}_{2}}(x)\ {{f}_{3}}(x)……}{{{g}_{1}}(x)\ {{g}_{2}}(x)\ {{g}_{3}}(x)……} \right)$$,

log y = log [f₁(x) f₂(x) f₃(x) …] – log [g₁(x) g₂(x) g₃(x) …].

Then it convenient to take the logarithm of the function first and then differentiate.

Note: write $$y={{\left[ f(x) \right]}^{g(x)}}={{e}^{g(x)\ln (f(x))}}$$ and differentiate easily or if y = [f (x)]g(x), then dy/dx = Differential of y treating f(x) as constant + Differential of y treating g(x) as constant.

Example:  Differentiate (log x)cosx with respect to x.

Solution: Let us consider y = (log x)cosx

Taking logarithm on both sides

log y = log (log x)cosx (∵ logam = m loga)

log y = cosx log (log x)

Differentiating both sides with respect to x

$$\frac{1}{y}.\frac{dy}{dx}=\log \left( \log x \right)\frac{d}{dx}\left( \cos x \right)+\cos x\frac{d}{dx}\left( \log \left( \log x \right) \right)$$,

$$=\log \left( \log x \right)\left( -\sin x \right)+\cos x\frac{1}{\log x}\frac{d}{dx}\left( \log x \right)$$,

$$=\log \left( \log x \right)\left( -\sin x \right)+\cos x\frac{1}{\log x}\left( \frac{1}{x} \right)$$,

$$\frac{dy}{dx}=y\left( -\sin x\log \left( \log x \right)+\frac{\cos x}{\log x\times x} \right)$$,

(since y = (logx)cosx)

$$={{\left( \log x \right)}^{\cos x}}\left( -\sin x\log \left( \log x \right)+\frac{\cos x}{\log x\times x} \right)$$.