Limits and Derivatives

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\underset{x\to 0}{lim}\frac{{\left(x+2\right)}^{\frac{1}{3}}-2^{\frac{1}{3}}}{x}\\ Putx+2=y\Rightarrow x=y-2\\ =\underset{y-2\to 0}{lim}\frac{y^{\frac{1}{3}}-2^{\frac{1}{3}}}{y-2}=\underset{y\to 2}{lim}\frac{y^{\frac{1}{3}}-2^{\frac{1}{3}}}{y-2}\\ =\frac{1}{3}.{\left(2\right)}^{\frac{1}{3}-1}=\frac{1}{3}.2^{\frac{-2}{3}}\left[\text{\hspace{0.17em}Using}\underset{x\to a}{lim}\frac{x^n-a^n}{x-a}=n.a^{n-1}\right]\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\underset{h\to 0}{lim}\frac{\sqrt[]{x+h}-\sqrt[]{x}}{h}\\ =\underset{h\to 0}{lim}\frac{\sqrt[]{x+h}-\sqrt[]{x}}{h}\times \frac{\sqrt[]{x+h}+\sqrt[]{x}}{\sqrt[]{x+h}+\sqrt[]{x}}\left[Rationalizingth\mathrm{e\; denominator}\right]\\ =\underset{h\to 0}{lim}\frac{x+h-x}{h\left[\sqrt[]{x+h}+\sqrt[]{x}\right]}=\underset{h\to 0}{lim}\frac{h}{h\left[\sqrt[]{x+h}+\sqrt[]{x}\right]}\\ =\underset{h\to 0}{lim}\frac{1}{\sqrt[]{x+h}+\sqrt[]{x}}\\ Taking\text{\hspace{0.17em}limit},wehave\\ \frac{1}{\sqrt[]{x}+\sqrt[]{x}}=\frac{1}{2\sqrt[]{x}}.\end{array}$

This is Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\underset{x\to \frac{1}{2}}{lim}\frac{4x^2-1}{2x-1}=\underset{x\to \frac{1}{2}}{lim}\frac{{\left(2x\right)}^2-{\left(1\right)}^2}{2x-1}=\underset{x\to \frac{1}{2}}{lim}\frac{\left(2x-1\right)\left(2x+1\right)}{2x-1}\\ =\underset{x\to \frac{1}{2}}{lim}2x+1\\ Taking,wehave\\ 2\times \frac{1}{2}+1=1+1=2\end{array}$

This is Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\underset{x\to 0}{lim}\frac{x^2-9}{x-3}=\underset{x\to 0}{lim}\frac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)}=\underset{x\to 0}{lim}x+3\\ Taking,wehave3+3=6.\end{array}$

18. $\underset{}{}$$\underset{x?0}{lim}\frac{ax+xcosx}{bsinx}=\underset{x?0}{lim}\frac{x(a+cosx)}{bsinx}$

$=\frac{1}{b}*\frac{(a+cos0)}{1}$

$=\frac{a+1}{b}$

73. Given, f (x) = $\frac{x}{si{n}^{n}x}$

So, f?(x) = $\frac{\left(si{n}^{x}x\frac{dx}{dx}\right)-\left(x\frac{d}{dx}si{n}^{x}x\right)}{{\left(si{n}^{x}x\right)}^2}$

= $\frac{si{n}^{x}x-x·n{(sinx)}^{x-1}\frac{d}{dx}sinx}{{(sinx)}^{2x}}$

$=\frac{{(sinx)}^n-x·x{(sinx)}^{n-1}·cosx}{{(sinx)}^{2n}}$

$=\frac{{(sinx)}^{n-1}[sinx-nx·cosx]}{{(sinx)}^{2n}}$

$=\frac{sinx-n·x·cosx}{{(sinx)}^{2n-(n-1)}}$

$=\frac{sinx-nxcosx}{{(sinx)}^{n+1}}$

72. Given, f (x) = (x + sec x) (x tan x)

So, f?(x) = (x + see x) $\frac{d}{dx}(x-tanx)+(x-tanx)\frac{d}{dx}(x+secx).$

$=(x+secx)\left(\frac{dx}{dx}-\frac{d}{dx}tanx\right)+(x-tanx)\left(\frac{dx}{dx}+\frac{d}{dx}secx\right)$

Let g(x) = tan x.

So, g?(x) =

$=\underset{h\to 0}{lim}\frac{1}{h}[tan(x+h)-tanx]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sin(x+h)}{cos(x+h)}-\frac{sinx}{cosx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sin(x+h)cosx-sinxcos(x+h)}{cos(x+h)cosx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sin(x+h-x)}{cos(x+h)cosx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{cos(x+h)cosx}\times \underset{h\to 0}{lim}\frac{sinh}{h}$

$=\frac{1}{co{s}^{2}x}=se{c}^{2}x$ . _____ (2)

And P(x) = see x.

$=\underset{h\to 0}{lim}\frac{sec(x+h)-secx}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{1}{cos(x+h)}-\frac{1}{cosx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{cosx-cos(x+h)}{cos(x+h)cosx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{-2sin\left(\frac{x+x+h}{2}\right)sin\left(\frac{x-(x+h)}{2}\right)}{cos(x+h)cosx}\right]$

$=\frac{sinx}{co{s}^{2}x}$ = tan x sec x. ____ (3)

   Putting (2) and (3) in (1) we get,

   f?(x) = (x + sec x) (1 - sec²x) + (x- tan x) (1 + tan x

71. Given, f (x) = $\frac{x}{1+tanx}$

$=\frac{x}{1+\frac{sinx}{cosx}}$

$=\frac{cosx·x}{cosx+sinx}$

$So,f^{\prime }(x)=\frac{(cosx+sinx)\frac{d}{dx}(x\cdot cosx)-\left(x\cdot cosx\right)\frac{d}{dx}\left(cosx+sinx\right)}{{(cosx+sinx)}^2}$

$=\frac{(cosx+sinx)(-xsinx+cosx)-\left(-x·sinxcosx+xco{s}^{2}x\right)}{{(cosx+sinx)}^2}$

$=\frac{-xcosxsinx+co{s}^{2}x-xsi{n}^{2}x+sinxcosx+xsinxcosx-xco{s}^{2}x}{{(cosx+sinx)}^2}$

$=\frac{co{s}^{2}x+sinxcosx-x\left(si{n}^{2}x+co{s}^{2}x\right)}{{(cosx+csinx)}^2}$

Dividing numerator and denominator by cos2x we get,

$=\frac{\frac{co{s}^{2}x}{co{s}^{2}x}+\frac{sinx}{cosx}·\frac{cosx}{cosx}-x\frac{1}{co{s}^{2}x}}{{\left(\frac{cosx}{cosx}+\frac{sinx}{cosx}\right)}^2}$

$=\frac{1+tanx-xse{c}^{2}x}{{(1+tanx)}^2}$

 70. Given, f (x) = $\frac{x^{2}cos\left(\frac{\pi }{4}\right)}{sinx}$

$=\frac{x^2}{sinx}\times cos\frac{\pi }{4}$

So, f? (x) =

$=cos\frac{\pi }{4}\left[\frac{2·xsinx-x^{2}cosx}{si{n}^{2}x}\right]$

$=xcos\frac{\pi }{4}\left[\frac{2sinx-xcosx}{si{n}^{2}x}\right].$

69. Given, f (x) = $\frac{4x+5sinx}{3x+7cosx}$

So,

$f^{\prime }(x)=\frac{(3x+7cosx)\frac{d}{dx}(4x+5sinx)-(4x+5sinx)\frac{d}{dx}(3x+7cosx)}{{(3x+7cosx)}^2}$

$=\frac{(3x+7cosx)(4x+5cosx)-(4x+5sinx)(3-7sinx)}{{(3x+7cosx)}^2}$

$\begin{array}{l}\underset{\_}{=x+xx+x+{}^{}x-x+xx-x+35si{n}^{2}x}\\ {\left(3x+7cosx\right)}^2\end{array}$

$=\frac{15xcosx+28cosx+28xsinx-15sinx+35\left(co{s}^{2}x+si{n}^{2}x\right)}{{(3x+7cosx)}^2}$

$=\frac{35+15xcosx+28cosx+28xsinx-15sinx}{{(3x+7cosx)}^2}$