Here, you will learn differentiability of a function and differentiability at a point and over an Interval.
Let’s begin –
Meaning of Derivative
The instantaneous rate of change of a function with respect to the dependent variable is called derivative. Let ‘f’ be a given function of one variable and let \(\Delta\)x denote a number (positive or negative) to be added to the number x. Let \(\Delta\)f denote the corresponding change of ‘f’ then \(\Delta\)f = f(x + \(\Delta\)x) – f(x).
\(\implies\) \(\Delta f\over {\Delta x}\) = \(f(x + \Delta x) – f(x)\over {\Delta x}\)
If \(\Delta f\over {\Delta x}\) approaches a limit as \(\Delta\)x approaches zero, this limit is the derivative of ‘f’ at the point x. The derivative of a function ‘f’ is a function; this function is denoted by symbols such as
\(f^{‘}(x)\), \(df\over {dx}\), \(d\over {dx}\)f(x) or \(df(x)\over {dx}\)
\(\implies\) \(df\over {dx}\) = \(\displaystyle{\lim_{\Delta x \to 0}}\)\(\Delta f\over {\Delta x}\) = \(\displaystyle{\lim_{\Delta x \to 0}}\) \(f(x + \Delta x) – f(x)\over {\Delta x}\)
The derivative evaluated at a point a, is written,
\(f^{‘}(a)\), \({df(x)\over {dx}}|_{x = a}\), \(f^{‘}(x)_{x = a}\)
Differentiable at x=a (Existence of Derivative at x = a)
(a) Right hand derivative
The right hand derivative of f(x) at x = a denoted by \(f(a^+)\) is defined as
\(f^{‘}(a^+)\) = \(\displaystyle{\lim_{h \to 0}}\) \(f(a + h) – f(a)\over h\), provided the limit exist and finite.(h > 0)
(b) Left hand derivative
The left hand derivative of f(x) at x = a denoted by \(f(a^-)\) is defined as
\(f^{‘}(a^-)\) = \(\displaystyle{\lim_{h \to 0}}\) \(f(a – h) – f(a)\over {-h}\), provided the limit exist and finite.(h > 0)
Hence f(x) is said to be derivable or differentiable at x = a. If \(f(a^+)\) = \(f^{‘}(a^-)\) = finite quantity and it is denoted by \(f^{‘}(a)\); where \(f^{‘}(a)\) = \(f^{‘}(a^+)\) = \(f^{‘}(a^-)\) & it is called derivative or differential coefficient of f(x) at x = a.
Differentiability of a function – Differentiable vs Continuous
If a function f(x) is derivable or differentiable at x = a, then f(x) is continuous at x = a.
Note :
(i) Differentiable \(\implies\) Continuous; Continuity \(\not\Rightarrow\) Differentiable; Not Differential \(\not\Rightarrow\) Not Continuous But Not Continuous \(\implies\) Not Differentiable
(ii) All polynomial, trigonometric, logarithmic and exponential function are continuous and differentiable in their domains.
(iii) If f(x) & g(x) are differentiable at x = a then the function f(x) + g(x), f(x) – g(x), f(x).g(x) will also be differentiable at x = a & g(a) \(\ne\) 0 then the function \(f(x)\over {g(x)}\) will also be differentiable at x = a.
Example : If f(x) = {[cos\(\pi\)x], x \(\leq\) 1 and 2{x} – 1, x > 1} comment on the derivability at x = 1, where [ ] denotes greatest integer function & { } denotes fractional part function.
Solution : For differentiability at x = 1, we determine, \(f^{‘}(1^-)\) and \(f^{‘}(1^+)\).
\(f^{‘}(1^-)\) = \(\displaystyle{\lim_{h \to 0}}\) \(f(1 – h) – f(1)\over {-h}\) = \(\displaystyle{\lim_{h \to 0}}\)\([cos(\pi – \pi h)] + 1\over {-h}\) = \(\displaystyle{\lim_{h \to 0}}\) \(-1 + 1\over {-h}\) = 0
\(f^{‘}(1^+)\) = \(\displaystyle{\lim_{h \to 0}}\) \(f(1 + h) – f(1)\over h\) = \(\displaystyle{\lim_{h \to 0}}\)\(2[1 + h] – 1 + 1\over h\) = \(\displaystyle{\lim_{h \to 0}}\)\(2h\over h\) = 2
Hence f(x) is not differentiable at x = 1.
Differentiability of a function over an Interval
(a) f(x) is said to be differentiable over an open interval (a, b) if it is differentiable at each and every point of the open interval (a, b).
(b) f(x) is said to be differentiable over the closed interval [a, b] if :
(i) f(x) is differentiable in (a, b) &
(ii) for the points a and b, \(f^{‘}(a^+)\) & \(f^{‘}(b^-)\) exist.
