If f is a real valued function and â€˜aâ€™ is any point in its domain for which f is defined then f(x) is said to be differentiable at the point x=a if the derivative f'(a) exists at every point in its domain. It is given by

\(f'(a) = \displaystyle{\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}}\)

Given that this limit exists and fâ€™(a) represents the derivative of f(x) at a. This is the first principle of derivative.

The domain of fâ€™(a) is defined by the existence of its limits. The derivative is also denoted as \(\frac{d}{dx}, f(x) \;\; or \;\; D(f(x))\)

If y = f(x) then derivative of f(x) is given as \(\frac{\mathrm{d} }{\mathrm{d} x}\)

This is known as derivative of y with respect to x.

Also, the derivative of a function f in x at x = a is given as:

\( \frac{\mathrm{d} }{\mathrm{d} x} f(x)|_{x = a}\)

Derivative of a function f(x) signifies the rate of change of the function f(x) with respect to x at a point lying in its domain. For a function to be differentiable at any point x = a in its domain, it must be continuous at that particular point but vice-versa is necessarily not always true.

The process of determining the derivative of a function is known as differentiation. It is clearly visible that the basic concept of derivative of a function is closely intertwined with limits. Therefore, it can be expected that the rules of derivatives are similar to that of limits. The following rules are a part of algebra of derivatives:

Consider f and g to be two real valued functions such that the differentiation of these functions is defined in a common domain. Then,

- Sum of derivatives of the functions f and g is equal to the derivative of their sum, i.e.,

\(\frac{\mathrm{d} }{\mathrm{d} x} \left [ f(x) + g(x) \right ] = \frac{\mathrm{d} }{\mathrm{d} x} f(x) + \frac{\mathrm{d} }{\mathrm{d} x} g(x)\)

Let u = f(x) and v = g(x), then (u + v)â€™ = uâ€™ + vâ€™

- Difference of derivatives of the functions f and g is equal to the derivative of difference of these functions, i.e.,

\(\frac{\mathrm{d} }{\mathrm{d} x} \left [ f(x) – g(x) \right ] = \frac{\mathrm{d} }{\mathrm{d} x} f(x) – \frac{\mathrm{d} }{\mathrm{d} x} g(x)\)

Let u = f(x) and v = g(x), then (u – v)â€™ = uâ€™ – vâ€™

- Derivative of the product of two functions f and g is given by the product rule as follows, i.e.,

\(\frac{\mathrm{d} }{\mathrm{d} x} \left [ f(x) . g(x) \right ] = g(x) \frac{\mathrm{d} }{\mathrm{d} x} f(x) + f(x)\frac{\mathrm{d} }{\mathrm{d} x} g(x)\)

Let u = f(x) and v = g(x), then the product rule can be restated as

(uv)â€™ = uâ€™v + uvâ€™

This is also known as Leibnitz rule for differentiating the products of functions

- The derivative of quotient of two functions f and g is given by the quotient rule (provided that the denominator is non- zero), i.e.,

\(\LARGE \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)} = \frac{g(x). \frac{\mathrm{d} f(x)}{\mathrm{d} x} – f(x). \frac{\mathrm{d} }{\mathrm{d} x}g(x)}{(g(x))^{2}}\)

Let u = f(x) and v = g(x), then the quotient rule can be restated as

\( \large \left ( \frac{u}{v} \right )’ = \frac{u’v – vu’}{v^{2}}\)

Chain Rule:

Whenever a quantity â€˜yâ€™ varies with another quantity â€˜xâ€™ such that y = f(x), then fâ€™(x) indicates the rate of change of y with respect to x (at x = x_{0}). Also, if two variables â€˜xâ€™ and â€˜yâ€™ are varying with respect to a third variable say â€˜tâ€™ then according to the chain rule, we have

\(\large \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right ) = \frac{\frac{\mathrm{d} y}{\mathrm{d} t}}{\frac{\mathrm{d} t}{\mathrm{d} x}},\;\; where \;\; \frac{\mathrm{d} t}{\mathrm{d} x}\neq 0\)

Let us look into some examples to have a better insight.

Letâ€™s Work Out- Example: Determine the rate of change of the volume of a sphere with respect to its radius â€˜râ€™ when r = 3 cm. Solution: We know the Volume of a Sphere is given as \(\frac{4}{3} \pi r^{3}\) Rate of change of Volume w.r.t. Radius is given as \(\frac{\mathrm{d} V}{\mathrm{d} r} = \frac{\mathrm{d} }{\mathrm{d} r}\left ( \frac{4}{3} \pi r^{3} \right ) = \frac{4}{3} \times \pi (3r^{2})\) \(\frac{\mathrm{d} V}{\mathrm{d} r} = 4 \pi r^{2}\) \(\frac{\mathrm{d} V}{\mathrm{d} r} \mid _{(at\;\; r=3)}= 4 \pi (3)^{2} = 36 \pi\) |

Learn more NCERT solutions forÂ Limits and Derivatives.

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