**Rule (I): **Differentiation of a constant function is zero i.e., d/dx (c) = 0

**Rule (II): **Let f(x) be a differentiable function and let c be a constant. Then c.f(x) is also differentiable such that \(\frac{d}{dx}\left\{ c.f\left( x \right) \right\}=c.\frac{d}{dx}\left( f\left( x \right) \right)\),

This is the derivative of a constant times a function is the constant times the derivative of the function.

**Rule (III): **If f(x) and g(x) are differentiable functions, then show that f(x) ± g(x) are also differentiable such that \(\frac{d}{dx}\left[ f\left( x \right)\pm g\left( x \right) \right]=\frac{d}{dx}f\left( x \right)\pm \frac{d}{dx}g\left( x \right)\),

That is the derivative of the sum or difference of two functions is the sum or difference of their derivatives.

**Rule (IV): **If f(x) and g(x) are two differentiable functions, then f(x).g(x) is also differentiable such that \(\frac{d}{dx}\left[ f\left( x \right).g\left( x \right) \right]=f\left( x \right)\frac{d}{dx}\left[ g\left( x \right) \right]+\frac{d}{dx}\left[ f\left( x \right) \right].g\left( x \right)\).

**Rule (V) (Quotient rule): **If f(x) and g(x) are two differentiable functions and g(x) ≠ 0 then f(x)/g(x) is also differentiable such that \(\frac{d}{dx}\left\{ \frac{f\left( x \right)}{g\left( x \right)} \right\}=\frac{g\left( x \right)\frac{d}{dx}\left[ f\left( x \right) \right]-f\left( x \right).\frac{d}{dx}\left[ g\left( x \right) \right]}{{{\left[ g\left( x \right) \right]}^{2}}}\).

**Relation between **dy/dx** and **dx/dy: Let x and y be two variables connected by a relation of the form f(x, y) = 0. Let Δx be a small change in x and let Δy be the corresponding change in y Then \(\frac{dy}{dx}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{\Delta y}{\Delta x}\) and \(\frac{dx}{dy}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{\Delta x}{\Delta y}\).Now,

\(\frac{\Delta y}{\Delta x}.\frac{\Delta x}{\Delta y}=1\),

⇒ \(\underset{\Delta x\to 0}{\mathop{\lim }}\,\left[ \frac{\Delta y}{\Delta x}.\frac{\Delta x}{\Delta y} \right]=1\),

⇒ \(\frac{dy}{dx}.\frac{dx}{dy}=1\) [∵ Δx → 0 ⇔ Δy → 0],

⇒ \(\frac{dy}{dx}.\frac{dx}{dy}=1\),

Hence, \(\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}\).

**Ex: **Differentiate f(x) = sinx.logx

**Solution: **Using product rule of differentiation

\(f’\left( x \right)=\frac{d}{dx}\left( \sin x \right).\log x+\sin x.\frac{d}{dx}\left( \log x \right)\),

= logx.cosx + sinx/x

**Ex: **Find the derivative of f(x) = logx/x²

**Solution: **Using quotient rule of differentiation

\(f’\left( x \right)=\frac{{{x}^{2}}\frac{d}{dx}\left( \log x \right)-\log x\frac{d}{dx}\left( {{x}^{2}} \right)}{{{\left( {{x}^{2}} \right)}^{2}}}\),

= \(\,\frac{{{x}^{2}}.\frac{1}{{{x}^{2}}}-\log x\left( 2x \right)}{{{x}^{4}}}\),

= \(\,\frac{x-2x\log x}{{{x}^{4}}}\),

= \(\,\frac{1-2\log x}{{{x}^{3}}}\).