Here, you will learn differentiability of a function and differentiability at a point and over an Interval.<\/p>\n
Let’s begin –<\/p>\n
Meaning of Derivative<\/strong><\/p>\n 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).<\/p>\n \\(\\implies\\) \\(\\Delta f\\over {\\Delta x}\\) = \\(f(x + \\Delta x) – f(x)\\over {\\Delta x}\\)<\/p>\n 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<\/p>\n \\(f^{‘}(x)\\), \\(df\\over {dx}\\), \\(d\\over {dx}\\)f(x) or \\(df(x)\\over {dx}\\)<\/p>\n<\/blockquote>\n \\(\\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}\\)<\/p>\n The derivative evaluated at a point a, is written,<\/p>\n \\(f^{‘}(a)\\), \\({df(x)\\over {dx}}|_{x = a}\\), \\(f^{‘}(x)_{x = a}\\)<\/p>\n<\/blockquote>\n (a) Right hand derivative<\/strong><\/p>\n The right hand derivative of f(x) at x = a denoted by \\(f(a^+)\\) is defined as<\/p>\n \\(f^{‘}(a^+)\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\(f(a + h) – f(a)\\over h\\), provided the limit exist and finite.(h > 0)<\/p>\n<\/blockquote>\n (b) Left hand derivative<\/strong><\/p>\n The left hand derivative of f(x) at x = a denoted by \\(f(a^-)\\) is defined as<\/p>\n \\(f^{‘}(a^-)\\) = \\(\\displaystyle{\\lim_{h \\to 0}}\\) \\(f(a – h) – f(a)\\over {-h}\\), provided the limit exist and finite.(h > 0)<\/p>\n<\/blockquote>\n 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.<\/p>\n If a function f(x) is derivable or differentiable at x = a, then f(x) is continuous at x = a.<\/p>\n Note :<\/strong><\/p>\n (i) Differentiable \\(\\implies\\) Continuous; Continuity \\(\\not\\Rightarrow\\) Differentiable; Not Differential \\(\\not\\Rightarrow\\) Not Continuous But Not Continuous \\(\\implies\\) Not Differentiable<\/p>\n (ii) All polynomial, trigonometric, logarithmic and exponential function are continuous and differentiable in their domains.<\/p>\n (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.<\/p>\n\n\n Example : <\/span>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.<\/p>\n Solution : <\/span>For differentiability at x = 1, we determine, \\(f^{‘}(1^-)\\) and \\(f^{‘}(1^+)\\). (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).<\/p>\n (b) f(x) is said to be differentiable over the closed interval [a, b] if :<\/p>\n (i) f(x) is differentiable in (a, b) &<\/p>\n (ii) for the points a and b, \\(f^{‘}(a^+)\\) & \\(f^{‘}(b^-)\\) exist.<\/p>\n<\/blockquote>\n\n\n\n
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Differentiable at x=a (Existence of Derivative at x = a)<\/h2>\n
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Differentiability of a function – Differentiable vs Continuous<\/h2>\n
\\(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.
<\/p>\n\n\nDifferentiability of a function over an Interval<\/h2>\n
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