NCERT Solutions · Class 11 Mathematics Limits and Derivatives

First principle. The derivative of \(\displaystyle f\) at \(\displaystyle x = a\) is\[f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}. \]Here \(\displaystyle f(x) = x^{2}-2 \) and \(\displaystyle a = 10 \), so \(\displaystyle f(10) = 100 - 2 = 98 \) and\[f'(10) = \lim_{h\to 0}\frac{\left[(10+h)^{2}-2\right] - 98}{h} = \lim_{h\to 0}\frac{100 + 20h + h^{2} - 2 - 98}{h}. \]\[= \lim_{h\to 0}\frac{20h + h^{2}}{h} = \lim_{h\to 0}\,(20 + h) = 20. \]Answer: \(\displaystyle 20\).

First principle. With \(\displaystyle f(x) = x \) and \(\displaystyle a = 1 \),\[f'(1) = \lim_{h\to 0}\frac{f(1+h)-f(1)}{h} = \lim_{h\to 0}\frac{(1+h)-1}{h} = \lim_{h\to 0}\frac{h}{h} = \lim_{h\to 0} 1 = 1. \](The cancellation is legal because \(\displaystyle h \neq 0\) while the limit is being taken.)Answer: \(\displaystyle 1\).

First principle. With \(\displaystyle f(x) = 99x \) and \(\displaystyle a = 100 \), so that \(\displaystyle f(100) = 9900 \),\[f'(100) = \lim_{h\to 0}\frac{99(100+h) - 9900}{h} = \lim_{h\to 0}\frac{9900 + 99h - 9900}{h} = \lim_{h\to 0}\frac{99h}{h} = 99. \]Answer: \(\displaystyle 99\).

First principle. For each function use\[f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}, \]simplifying the difference quotient until the \(\displaystyle h\) in the denominator cancels.(i) \(\displaystyle f(x) = x^{3}-27 \). Since \(\displaystyle (x+h)^{3} = x^{3}+3x^{2}h+3xh^{2}+h^{3} \),\[\frac{f(x+h)-f(x)}{h} = \frac{\left(x^{3}+3x^{2}h+3xh^{2}+h^{3}-27\right)-\left(x^{3}-27\right)}{h} = \frac{3x^{2}h+3xh^{2}+h^{3}}{h} = 3x^{2}+3xh+h^{2}. \]Letting \(\displaystyle h \to 0 \): \(\displaystyle f'(x) = 3x^{2} \).(ii) \(\displaystyle f(x) = (x-1)(x-2) = x^{2}-3x+2 \).\[\frac{f(x+h)-f(x)}{h} = \frac{\left[(x+h)^{2}-3(x+h)+2\right]-\left[x^{2}-3x+2\right]}{h} = \frac{2xh+h^{2}-3h}{h} = 2x+h-3. \]Letting \(\displaystyle h \to 0 \): \(\displaystyle f'(x) = 2x-3 \).(iii) \(\displaystyle f(x) = \dfrac{1}{x^{2}} \), \(\displaystyle x \neq 0\).\[\frac{f(x+h)-f(x)}{h} = \frac{1}{h}\left[\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}\right] = \frac{1}{h}\cdot\frac{x^{2}-(x+h)^{2}}{x^{2}(x+h)^{2}} = \frac{1}{h}\cdot\frac{-2xh-h^{2}}{x^{2}(x+h)^{2}}. \]\[= \frac{-(2x+h)}{x^{2}(x+h)^{2}} \; \xrightarrow[h\to 0]{} \; \frac{-2x}{x^{2}\cdot x^{2}} = -\frac{2}{x^{3}}. \]So \(\displaystyle f'(x) = -\dfrac{2}{x^{3}} \).(iv) \(\displaystyle f(x) = \dfrac{x+1}{x-1} \), \(\displaystyle x \neq 1\). Combine the two fractions first:\[f(x+h)-f(x) = \frac{x+h+1}{x+h-1}-\frac{x+1}{x-1} = \frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x+h-1)(x-1)}. \]Expanding the numerator, \(\displaystyle (x+h+1)(x-1) = x^{2}+hx-h-1 \) and \(\displaystyle (x+1)(x+h-1) = x^{2}+hx+h-1 \), so their difference is \(\displaystyle -2h\). Hence\[\frac{f(x+h)-f(x)}{h} = \frac{-2h}{h\,(x+h-1)(x-1)} = \frac{-2}{(x+h-1)(x-1)} \; \xrightarrow[h\to 0]{} \; \frac{-2}{(x-1)^{2}}. \]So \(\displaystyle f'(x) = -\dfrac{2}{(x-1)^{2}} \).Answer: (i) \(\displaystyle 3x^{2}\) (ii) \(\displaystyle 2x-3\) (iii) \(\displaystyle -\dfrac{2}{x^{3}}\) (iv) \(\displaystyle -\dfrac{2}{(x-1)^{2}}\).

Term-by-term differentiation.The derivative of a sum is the sum of the derivatives, and \(\displaystyle \dfrac{d}{dx}x^{k}=kx^{k-1} \).The general term of \(\displaystyle f \) is \(\displaystyle \dfrac{x^{k}}{k} \) for \(\displaystyle k=1,2,\ldots,100 \), and\[\frac{d}{dx}\left(\frac{x^{k}}{k}\right)=\frac{1}{k}\cdot kx^{k-1}=x^{k-1}. \]The last term \(\displaystyle 1 \) is a constant, so it differentiates to \(\displaystyle 0 \). Therefore\[f'(x)=x^{99}+x^{98}+\ldots+x^{2}+x+1, \]a sum of exactly \(\displaystyle 100 \) terms.Value at \(\displaystyle x=1 \). Every one of the \(\displaystyle 100 \) terms equals \(\displaystyle 1 \):\[f'(1)=\underbrace{1+1+\ldots+1}_{100\ \text{terms}}=100. \]Value at \(\displaystyle x=0 \). Every term containing a positive power of \(\displaystyle x \) vanishes, leaving only the constant term \(\displaystyle 1 \):\[f'(0)=0+0+\ldots+0+1=1. \]Hence \(\displaystyle 100\,f'(0)=100\times 1=100=f'(1) \).Therefore \(\displaystyle f'(1)=100\,f'(0) \), as required.

Term-by-term differentiation.Here \(\displaystyle a \) is a fixed real number, so \(\displaystyle a,\,a^{2},\ldots,a^{n} \) are all constants; only the powers of \(\displaystyle x \) are differentiated. Using \(\displaystyle \dfrac{d}{dx}x^{m}=mx^{m-1} \) and the fact that the derivative of a constant is \(\displaystyle 0 \):Let \(\displaystyle y=x^{n}+ax^{n-1}+a^{2}x^{n-2}+\ldots+a^{n-1}x+a^{n} \). Its general term is \(\displaystyle a^{k}x^{n-k} \) for \(\displaystyle k=0,1,\ldots,n \), which differentiates to \(\displaystyle (n-k)a^{k}x^{n-k-1} \). The last term \(\displaystyle a^{n} \) (the case \(\displaystyle k=n \)) is constant and contributes nothing.\[\frac{dy}{dx}=nx^{n-1}+(n-1)ax^{n-2}+(n-2)a^{2}x^{n-3}+\ldots+2a^{n-2}x+a^{n-1}. \]\(\displaystyle \dfrac{dy}{dx}=\displaystyle\sum_{k=0}^{n-1}(n-k)\,a^{k}x^{\,n-k-1}=nx^{n-1}+(n-1)ax^{n-2}+(n-2)a^{2}x^{n-3}+\ldots+2a^{n-2}x+a^{n-1} \)

Product rule and quotient rule.For differentiable \(\displaystyle u,v \): \(\displaystyle (uv)'=u'v+uv' \) and \(\displaystyle \left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}} \) (where \(\displaystyle v\neq 0 \)).(i) \(\displaystyle (x-a)(x-b) \). Expanding first is quickest:\[(x-a)(x-b)=x^{2}-(a+b)x+ab \quad\Rightarrow\quad \frac{d}{dx}\big[(x-a)(x-b)\big]=2x-(a+b). \](The product rule gives the same thing: \(\displaystyle 1\cdot(x-b)+(x-a)\cdot 1=2x-a-b \).)(ii) \(\displaystyle (ax^{2}+b)^{2} \). Expand, then differentiate:\[(ax^{2}+b)^{2}=a^{2}x^{4}+2abx^{2}+b^{2} \] \[\frac{d}{dx}\big[(ax^{2}+b)^{2}\big]=4a^{2}x^{3}+4abx=4ax\,(ax^{2}+b). \](iii) \(\displaystyle \dfrac{x-a}{x-b} \). Take \(\displaystyle u=x-a,\ v=x-b \), so \(\displaystyle u'=1,\ v'=1 \) and \(\displaystyle x\neq b \):\[\frac{d}{dx}\left(\frac{x-a}{x-b}\right)=\frac{1\cdot(x-b)-(x-a)\cdot 1}{(x-b)^{2}}=\frac{x-b-x+a}{(x-b)^{2}}=\frac{a-b}{(x-b)^{2}}. \](i) \(\displaystyle 2x-a-b \) (ii) \(\displaystyle 4ax(ax^{2}+b)=4a^{2}x^{3}+4abx \) (iii) \(\displaystyle \dfrac{a-b}{(x-b)^{2}} \)

Quotient rule.Let \(\displaystyle u=x^{n}-a^{n} \) and \(\displaystyle v=x-a \), so that \(\displaystyle u'=nx^{n-1} \) and \(\displaystyle v'=1 \) (\(\displaystyle a^{n} \) is a constant). For \(\displaystyle x\neq a \),\[\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^{2}}=\frac{nx^{n-1}(x-a)-(x^{n}-a^{n})\cdot 1}{(x-a)^{2}}. \]Expanding the numerator:\[nx^{n}-nax^{n-1}-x^{n}+a^{n}=(n-1)x^{n}-nax^{n-1}+a^{n}. \]A tidier route. Since \(\displaystyle x^{n}-a^{n}=(x-a)\left(x^{n-1}+ax^{n-2}+\ldots+a^{n-1}\right) \), for \(\displaystyle x\neq a \) the function is simply the polynomial \(\displaystyle x^{n-1}+ax^{n-2}+\ldots+a^{n-1} \). Differentiating that term by term gives the same derivative in the form \(\displaystyle (n-1)x^{n-2}+(n-2)ax^{n-3}+\ldots+a^{n-2} \).\(\displaystyle \dfrac{d}{dx}\left(\dfrac{x^{n}-a^{n}}{x-a}\right)=\dfrac{(n-1)x^{n}-nax^{n-1}+a^{n}}{(x-a)^{2}} \), equivalently \(\displaystyle (n-1)x^{n-2}+(n-2)ax^{n-3}+\ldots+a^{n-2} \)

Power rule, product rule and quotient rule.Throughout, \(\displaystyle \dfrac{d}{dx}x^{m}=mx^{m-1} \), the derivative of a constant is \(\displaystyle 0 \), and \(\displaystyle (uv)'=u'v+uv' \), \(\displaystyle \left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}} \).(i) \(\displaystyle 2x-\dfrac{3}{4} \). \[\frac{d}{dx}\left(2x-\frac{3}{4}\right)=2-0=2. \](ii) \(\displaystyle (5x^{3}+3x-1)(x-1) \). By the product rule with \(\displaystyle u=5x^{3}+3x-1 \), \(\displaystyle v=x-1 \): \[=(15x^{2}+3)(x-1)+(5x^{3}+3x-1)(1) \] \[=15x^{3}-15x^{2}+3x-3+5x^{3}+3x-1=20x^{3}-15x^{2}+6x-4. \](iii) \(\displaystyle x^{-3}(5+3x) \). Multiply out first: \(\displaystyle 5x^{-3}+3x^{-2} \). \[\frac{d}{dx}\left(5x^{-3}+3x^{-2}\right)=-15x^{-4}-6x^{-3}=-\frac{3(5+2x)}{x^{4}}. \](iv) \(\displaystyle x^{5}(3-6x^{-9}) \). Multiply out: \(\displaystyle 3x^{5}-6x^{-4} \). \[\frac{d}{dx}\left(3x^{5}-6x^{-4}\right)=15x^{4}+24x^{-5}=15x^{4}+\frac{24}{x^{5}}. \](v) \(\displaystyle x^{-4}(3-4x^{-5}) \). Multiply out: \(\displaystyle 3x^{-4}-4x^{-9} \). \[\frac{d}{dx}\left(3x^{-4}-4x^{-9}\right)=-12x^{-5}+36x^{-10}=-\frac{12}{x^{5}}+\frac{36}{x^{10}}. \](vi) \(\displaystyle \dfrac{2}{x+1}-\dfrac{x^{2}}{3x-1} \). Differentiate each piece by the quotient rule. \[\frac{d}{dx}\left(\frac{2}{x+1}\right)=\frac{0\cdot(x+1)-2\cdot 1}{(x+1)^{2}}=\frac{-2}{(x+1)^{2}} \] \[\frac{d}{dx}\left(\frac{x^{2}}{3x-1}\right)=\frac{2x(3x-1)-x^{2}\cdot 3}{(3x-1)^{2}}=\frac{6x^{2}-2x-3x^{2}}{(3x-1)^{2}}=\frac{3x^{2}-2x}{(3x-1)^{2}} \] Subtracting, \[=-\frac{2}{(x+1)^{2}}-\frac{3x^{2}-2x}{(3x-1)^{2}}. \](i) \(\displaystyle 2 \) (ii) \(\displaystyle 20x^{3}-15x^{2}+6x-4 \) (iii) \(\displaystyle -\dfrac{15}{x^{4}}-\dfrac{6}{x^{3}} \) (iv) \(\displaystyle 15x^{4}+\dfrac{24}{x^{5}} \) (v) \(\displaystyle -\dfrac{12}{x^{5}}+\dfrac{36}{x^{10}} \) (vi) \(\displaystyle -\dfrac{2}{(x+1)^{2}}-\dfrac{3x^{2}-2x}{(3x-1)^{2}} \)

First principle.By definition, \(\displaystyle f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \). With \(\displaystyle f(x)=\cos x \),\[f'(x)=\lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h}. \]Use the transformation \(\displaystyle \cos C-\cos D=-2\sin\dfrac{C+D}{2}\,\sin\dfrac{C-D}{2} \) with \(\displaystyle C=x+h \) and \(\displaystyle D=x \), so \(\displaystyle \dfrac{C+D}{2}=x+\dfrac{h}{2} \) and \(\displaystyle \dfrac{C-D}{2}=\dfrac{h}{2} \):\[f'(x)=\lim_{h\to 0}\frac{-2\sin\left(x+\dfrac{h}{2}\right)\sin\dfrac{h}{2}}{h} =\lim_{h\to 0}\left[-\sin\left(x+\frac{h}{2}\right)\cdot\frac{\sin\dfrac{h}{2}}{\dfrac{h}{2}}\right]. \]As \(\displaystyle h\to 0 \), \(\displaystyle \dfrac{h}{2}\to 0 \), so by the standard limit \(\displaystyle \displaystyle\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1 \) the second factor tends to \(\displaystyle 1 \), while \(\displaystyle \sin\left(x+\dfrac{h}{2}\right)\to\sin x \) by continuity of the sine function.\[f'(x)=-\sin x\cdot 1=-\sin x. \]\(\displaystyle \dfrac{d}{dx}(\cos x)=-\sin x \)

Standard trigonometric derivatives with the product and quotient rules.We start from \(\displaystyle \dfrac{d}{dx}(\sin x)=\cos x \) and \(\displaystyle \dfrac{d}{dx}(\cos x)=-\sin x \), and use \(\displaystyle (uv)'=u'v+uv' \), \(\displaystyle \left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}} \).(i) \(\displaystyle \sin x\cos x \). By the product rule, \[=\cos x\cdot\cos x+\sin x\cdot(-\sin x)=\cos^{2}x-\sin^{2}x=\cos 2x. \](ii) \(\displaystyle \sec x=\dfrac{1}{\cos x} \). By the quotient rule (\(\displaystyle \cos x\neq 0 \)), \[=\frac{0\cdot\cos x-1\cdot(-\sin x)}{\cos^{2}x}=\frac{\sin x}{\cos^{2}x}=\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}=\sec x\tan x. \](iii) \(\displaystyle 5\sec x+4\cos x \). Using (ii) and \(\displaystyle (\cos x)'=-\sin x \), \[=5\sec x\tan x-4\sin x. \](iv) \(\displaystyle \operatorname{cosec} x=\dfrac{1}{\sin x} \). By the quotient rule (\(\displaystyle \sin x\neq 0 \)), \[=\frac{0\cdot\sin x-1\cdot\cos x}{\sin^{2}x}=-\frac{\cos x}{\sin^{2}x}=-\operatorname{cosec} x\cot x. \](v) \(\displaystyle 3\cot x+5\operatorname{cosec} x \). First \(\displaystyle \cot x=\dfrac{\cos x}{\sin x} \): \[\frac{d}{dx}(\cot x)=\frac{-\sin x\cdot\sin x-\cos x\cdot\cos x}{\sin^{2}x}=\frac{-(\sin^{2}x+\cos^{2}x)}{\sin^{2}x}=-\operatorname{cosec}^{2}x. \] With (iv), the derivative is \[-3\operatorname{cosec}^{2}x-5\operatorname{cosec} x\cot x. \](vi) \(\displaystyle 5\sin x-6\cos x+7 \). Differentiating term by term, \[=5\cos x-6(-\sin x)+0=5\cos x+6\sin x. \](vii) \(\displaystyle 2\tan x-7\sec x \). First \(\displaystyle \tan x=\dfrac{\sin x}{\cos x} \): \[\frac{d}{dx}(\tan x)=\frac{\cos x\cdot\cos x-\sin x\cdot(-\sin x)}{\cos^{2}x}=\frac{1}{\cos^{2}x}=\sec^{2}x. \] With (ii), the derivative is \[2\sec^{2}x-7\sec x\tan x. \](i) \(\displaystyle \cos 2x \) (ii) \(\displaystyle \sec x\tan x \) (iii) \(\displaystyle 5\sec x\tan x-4\sin x \) (iv) \(\displaystyle -\operatorname{cosec} x\cot x \) (v) \(\displaystyle -3\operatorname{cosec}^{2}x-5\operatorname{cosec} x\cot x \) (vi) \(\displaystyle 5\cos x+6\sin x \) (vii) \(\displaystyle 2\sec^{2}x-7\sec x\tan x \)