NCERT Solutions · Class 11 Mathematics Limits and Derivatives

First principle.Throughout, \(\displaystyle f'(x)=\displaystyle\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \).(i) \(\displaystyle f(x)=-x \). \[f'(x)=\lim_{h\to 0}\frac{-(x+h)-(-x)}{h}=\lim_{h\to 0}\frac{-x-h+x}{h}=\lim_{h\to 0}\frac{-h}{h}=-1. \](ii) \(\displaystyle f(x)=(-x)^{-1}=-\dfrac{1}{x} \) (for \(\displaystyle x\neq 0 \)). \[f'(x)=\lim_{h\to 0}\frac{1}{h}\left(-\frac{1}{x+h}+\frac{1}{x}\right)=\lim_{h\to 0}\frac{1}{h}\cdot\frac{-x+(x+h)}{x(x+h)}=\lim_{h\to 0}\frac{1}{h}\cdot\frac{h}{x(x+h)} \] \[=\lim_{h\to 0}\frac{1}{x(x+h)}=\frac{1}{x^{2}}. \](iii) \(\displaystyle f(x)=\sin(x+1) \). Use \(\displaystyle \sin C-\sin D=2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2} \) with \(\displaystyle C=x+h+1 \), \(\displaystyle D=x+1 \): \[f'(x)=\lim_{h\to 0}\frac{\sin(x+h+1)-\sin(x+1)}{h}=\lim_{h\to 0}\frac{2\cos\left(x+1+\dfrac{h}{2}\right)\sin\dfrac{h}{2}}{h} \] \[=\lim_{h\to 0}\left[\cos\left(x+1+\frac{h}{2}\right)\cdot\frac{\sin\dfrac{h}{2}}{\dfrac{h}{2}}\right]=\cos(x+1)\cdot 1=\cos(x+1). \](iv) \(\displaystyle f(x)=\cos\left(x-\dfrac{\pi}{8}\right) \). Use \(\displaystyle \cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2} \) with \(\displaystyle C=x+h-\dfrac{\pi}{8} \), \(\displaystyle D=x-\dfrac{\pi}{8} \): \[f'(x)=\lim_{h\to 0}\frac{-2\sin\left(x-\dfrac{\pi}{8}+\dfrac{h}{2}\right)\sin\dfrac{h}{2}}{h} =\lim_{h\to 0}\left[-\sin\left(x-\frac{\pi}{8}+\frac{h}{2}\right)\cdot\frac{\sin\dfrac{h}{2}}{\dfrac{h}{2}}\right] \] \[=-\sin\left(x-\frac{\pi}{8}\right). \](Each limit uses \(\displaystyle \displaystyle\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1 \) with \(\displaystyle \theta=\dfrac{h}{2} \).)(i) \(\displaystyle -1 \) (ii) \(\displaystyle \dfrac{1}{x^{2}} \) (iii) \(\displaystyle \cos(x+1) \) (iv) \(\displaystyle -\sin\left(x-\dfrac{\pi}{8}\right) \)

Sum rule.Here \(\displaystyle a \) is a fixed constant, so \(\displaystyle \dfrac{d}{dx}(a)=0 \), and \(\displaystyle \dfrac{d}{dx}(x)=1 \).\[\frac{d}{dx}(x+a)=\frac{d}{dx}(x)+\frac{d}{dx}(a)=1+0=1. \]\(\displaystyle \dfrac{d}{dx}(x+a)=1 \)

Expand first, then the power rule.The product expands to a sum of powers of \(\displaystyle x \), which avoids the quotient rule entirely. For \(\displaystyle x\neq 0 \),\[(px+q)\left(\frac{r}{x}+s\right)=px\cdot\frac{r}{x}+px\cdot s+q\cdot\frac{r}{x}+qs=pr+psx+qrx^{-1}+qs. \]Here \(\displaystyle pr \) and \(\displaystyle qs \) are constants, so they differentiate to \(\displaystyle 0 \); and \(\displaystyle \dfrac{d}{dx}x^{-1}=-x^{-2} \):\[\frac{d}{dx}\left[(px+q)\left(\frac{r}{x}+s\right)\right]=0+ps-qrx^{-2}+0=ps-\frac{qr}{x^{2}}. \]\(\displaystyle ps-\dfrac{qr}{x^{2}} \)

Product rule.Write \(\displaystyle u=ax+b \) and \(\displaystyle v=(cx+d)^{2} \), so \(\displaystyle (uv)'=u'v+uv' \).Here \(\displaystyle u'=a \). For \(\displaystyle v \), expand before differentiating: \[v=(cx+d)^{2}=c^{2}x^{2}+2cdx+d^{2}\quad\Rightarrow\quad v'=2c^{2}x+2cd=2c(cx+d). \]Therefore \[\frac{d}{dx}\left[(ax+b)(cx+d)^{2}\right]=a(cx+d)^{2}+2c(ax+b)(cx+d). \]Taking out the common factor \(\displaystyle (cx+d) \): \[=(cx+d)\big[a(cx+d)+2c(ax+b)\big]=(cx+d)\big(acx+ad+2acx+2bc\big)=(cx+d)(3acx+ad+2bc). \]\(\displaystyle a(cx+d)^{2}+2c(ax+b)(cx+d)=(cx+d)(3acx+ad+2bc) \)

Quotient rule.If \(\displaystyle u=ax+b \) and \(\displaystyle v=cx+d \) then \(\displaystyle u'=a \), \(\displaystyle v'=c \), and \(\displaystyle \left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}} \) for \(\displaystyle cx+d\neq 0 \):\[\frac{d}{dx}\left(\frac{ax+b}{cx+d}\right)=\frac{a(cx+d)-(ax+b)c}{(cx+d)^{2}}=\frac{acx+ad-acx-bc}{(cx+d)^{2}}=\frac{ad-bc}{(cx+d)^{2}}. \]Notice the \(\displaystyle x \) terms cancel, so the derivative has a constant numerator.\(\displaystyle \dfrac{ad-bc}{(cx+d)^{2}} \)

Simplify first, then the quotient rule.For \(\displaystyle x\neq 0 \), multiply numerator and denominator by \(\displaystyle x \):\[\frac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=\frac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}}=\frac{x+1}{x-1}\qquad (x\neq 0,\ x\neq 1). \]Now apply \(\displaystyle \left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}} \) with \(\displaystyle u=x+1 \), \(\displaystyle v=x-1 \), so \(\displaystyle u'=v'=1 \):\[\frac{d}{dx}\left(\frac{x+1}{x-1}\right)=\frac{1\cdot(x-1)-(x+1)\cdot 1}{(x-1)^{2}}=\frac{x-1-x-1}{(x-1)^{2}}=\frac{-2}{(x-1)^{2}}. \]\(\displaystyle -\dfrac{2}{(x-1)^{2}} \)

Quotient rule.Take \(\displaystyle u=1 \) and \(\displaystyle v=ax^{2}+bx+c \), so \(\displaystyle u'=0 \) and \(\displaystyle v'=2ax+b \). For \(\displaystyle ax^{2}+bx+c\neq 0 \),\[\frac{d}{dx}\left(\frac{1}{ax^{2}+bx+c}\right)=\frac{0\cdot(ax^{2}+bx+c)-1\cdot(2ax+b)}{(ax^{2}+bx+c)^{2}}=\frac{-(2ax+b)}{(ax^{2}+bx+c)^{2}}. \]\(\displaystyle -\dfrac{2ax+b}{\left(ax^{2}+bx+c\right)^{2}} \)

Quotient rule.Take \(\displaystyle u=ax+b \) and \(\displaystyle v=px^{2}+qx+r \), so \(\displaystyle u'=a \) and \(\displaystyle v'=2px+q \). For \(\displaystyle px^{2}+qx+r\neq 0 \),\[\frac{d}{dx}\left(\frac{ax+b}{px^{2}+qx+r}\right)=\frac{a\left(px^{2}+qx+r\right)-(ax+b)(2px+q)}{\left(px^{2}+qx+r\right)^{2}}. \]Expanding the numerator: \[\big(apx^{2}+aqx+ar\big)-\big(2apx^{2}+aqx+2bpx+bq\big)=-apx^{2}-2bpx+ar-bq. \]\(\displaystyle \dfrac{-apx^{2}-2bpx+ar-bq}{\left(px^{2}+qx+r\right)^{2}} \)

Quotient rule.Take \(\displaystyle u=px^{2}+qx+r \) and \(\displaystyle v=ax+b \), so \(\displaystyle u'=2px+q \) and \(\displaystyle v'=a \). For \(\displaystyle ax+b\neq 0 \),\[\frac{d}{dx}\left(\frac{px^{2}+qx+r}{ax+b}\right)=\frac{(2px+q)(ax+b)-\left(px^{2}+qx+r\right)a}{(ax+b)^{2}}. \]Expanding the numerator: \[\big(2apx^{2}+2bpx+aqx+bq\big)-\big(apx^{2}+aqx+ar\big)=apx^{2}+2bpx+bq-ar. \]\(\displaystyle \dfrac{apx^{2}+2bpx+bq-ar}{(ax+b)^{2}} \)

Rewrite with negative indices, then differentiate term by term.For \(\displaystyle x\neq 0 \), \[\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x=ax^{-4}-bx^{-2}+\cos x. \]Using \(\displaystyle \dfrac{d}{dx}x^{m}=mx^{m-1} \) and \(\displaystyle \dfrac{d}{dx}(\cos x)=-\sin x \):\[\frac{d}{dx}\left(ax^{-4}-bx^{-2}+\cos x\right)=-4ax^{-5}+2bx^{-3}-\sin x. \]\(\displaystyle -\dfrac{4a}{x^{5}}+\dfrac{2b}{x^{3}}-\sin x \)