Derivative of A Function- Calculus

If f is a real valued function and ‘a’ 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 f’(a) represents the derivative of f(x) at a. This is the first principle of derivative.
The domain of f’(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 \(\frac{\mathrm{d} }{\mathrm{d} x}\) or y’.

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:

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

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)’ = u’ + v’

  • 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)’ = u’ – v’

  • 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)’ = u’v + uv’

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 ‘y’ varies with another quantity ‘x’ such that y = f(x), then f’(x) indicates the rate of change of y with respect to x (at x = x0). Also, if two variables ‘x’ and ‘y’ are varying with respect to a third variable say ‘t’ 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.

Let’s Work Out-

Example: Determine the rate of change of the volume of a sphere with respect to its radius ‘r’ 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 for Limits and Derivatives.


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Whole numbers consists of all the natural numbers and 0.