Miscellaneous Example (Exercise) - Chapter 13 - Limits & Derivatives - Ncert Solutions class 11 - Maths

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Chapter 12: Limits & Derivatives - NCERT Solutions for Class 11 Maths

Example 1

Find the derivative of $f$ from the first principle, where $f$ is given by
(i) $f(x)=\frac{2 x+3}{x-2}$
(ii) $f(x)=x+\frac{1}{x}$

Solution

(i) Note that function is not defined at $x=2$. But, we have
$
f^{\prime}(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{\frac{2(x+h)+3}{x+h-2}-\frac{2 x+3}{x-2}}{h}
$

$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{(2 x+2 h+3)(x-2)-(2 x+3)(x+h-2)}{h(x-2)(x+h-2)} \\
& =\lim _{h \rightarrow 0} \frac{(2 x+3)(x-2)+2 h(x-2)-(2 x+3)(x-2)-h(2 x+3)}{h(x-2)(x+h-2)} \\
& =\lim _{h \rightarrow 0} \frac{-7}{(x-2)(x+h-2)}=-\frac{7}{(x-2)^2}
\end{aligned}
$

Again, note that the function $f^{\prime}$ is also not defined at $x=2$.
(ii) The function is not defined at $x=0$. But, we have
$
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{\left(x+h+\frac{1}{x+h}\right)-\left(x+\frac{1}{x}\right)}{h} \\
& =\lim _{h \rightarrow 0} \frac{1}{h}\left[h+\frac{1}{x+h}-\frac{1}{x}\right] \\
& =\lim _{h \rightarrow 0} \frac{1}{h}\left[h+\frac{x-x-h}{x(x+h)}\right]=\lim _{h \rightarrow 0} \frac{1}{h}\left[h\left(1-\frac{1}{x(x+h)}\right)\right] \\
& =\lim _{h \rightarrow 0}\left[1-\frac{1}{x(x+h)}\right]=1-\frac{1}{x^2}
\end{aligned}
$

Again, note that the function $f^{\prime}$ is not defined at $x=0$.

Example 2

 Find the derivative of $f(x)$ from the first principle, where $f(x)$ is
(i) $\sin x+\cos x$
(ii) $x \sin x$

Solution

(i) we have $f^{\prime}(x)=\frac{f(x+h)-f(x)}{h}$
$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\sin (x+h)+\cos (x+h)-\sin x-\cos x}{h} \\
& =\lim _{h \rightarrow 0} \frac{\sin x \cos h+\cos x \sin h+\cos x \cos h-\sin x \sin h-\sin x-\cos x}{h}
\end{aligned}
$

$
\begin{aligned}
& =\lim _{h \rightarrow 0} \frac{\sin h(\cos x-\sin x)+\sin x(\cos h-1)+\cos x(\cos h-1)}{h} \\
& =\lim _{h \rightarrow 0} \frac{\sin h}{h}(\cos x-\sin x)+\lim _{h \rightarrow 0} \sin x \frac{(\cos h-1)}{h}+\lim _{h \rightarrow 0} \cos x \frac{(\cos h-1)}{h} \\
& =\cos x-\sin x
\end{aligned}
$
(ii)
$
\begin{aligned}
f^{\prime}(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0} \frac{(x+h) \sin (x+h)-x \sin x}{h} \\
& =\lim _{h \rightarrow 0} \frac{(x+h)(\sin x \cos h+\sin h \cos x)-x \sin x}{h} \\
& =\lim _{h \rightarrow 0} \frac{x \sin x(\cos h-1)+x \cos x \sin h+h(\sin x \cos h+\sin h \cos x)}{h} \\
= & \lim _{h \rightarrow 0} \frac{x \sin x(\cos h-1)}{h}+\lim _{h \rightarrow 0} x \cos x \frac{\sin h}{h}+\lim _{h \rightarrow 0}(\sin x \cos h+\sin h \cos x) \\
& =x \cos x+\sin x
\end{aligned}
$

Example 3

Compute derivative of
(i) $f(x)=\sin 2 x$
(ii) $g(x)=\cot x$

Solution

(i) Recall the trigonometric formula $\sin 2 x=2 \sin x \cos x$. Thus
$
\begin{aligned}
\frac{d f(x)}{d x} & =\frac{d}{d x}(2 \sin x \cos x)=2 \frac{d}{d x}(\sin x \cos x) \\
& =2\left[(\sin x)^{\prime} \cos x+\sin x(\cos x)^{\prime}\right]
\end{aligned}
$

$
\begin{aligned}
& =2[(\cos x) \cos x+\sin x(-\sin x)] \\
& =2\left(\cos ^2 x-\sin ^2 x\right)
\end{aligned}
$
(ii) By definition, $\mathrm{g}(x)=\cot x=\frac{\cos x}{\sin x}$. We use the quotient rule on this function wherever it is defined. $\frac{d g}{d x}=\frac{d}{d x}(\cot x)=\frac{d}{d x}\left(\frac{\cos x}{\sin x}\right)$

$
\begin{aligned}
& =\frac{(\cos x)^{\prime}(\sin x)-(\cos x)(\sin x)^{\prime}}{(\sin x)^2} \\
& =\frac{(-\sin x)(\sin x)-(\cos x)(\cos x)}{(\sin x)^2} \\
& =-\frac{\sin ^2 x+\cos ^2 x}{\sin ^2 x}=-\operatorname{cosec}^2 x
\end{aligned}
$

Alternatively, this may be computed by noting that $\cot x=\frac{1}{\tan x}$. Here, we use the fact that the derivative of $\tan x$ is $\sec ^2 x$ which we saw in Example 17 and also that the derivative of the constant function is 0 .
$
\begin{aligned}
\frac{d g}{d x} & =\frac{d}{d x}(\cot x)=\frac{d}{d x}\left(\frac{1}{\tan x}\right) \\
& =\frac{(1)^{\prime}(\tan x)-(1)(\tan x)^{\prime}}{(\tan x)^2} \\
& =\frac{(0)(\tan x)-(\sec x)^2}{(\tan x)^2} \\
& =\frac{-\sec ^2 x}{\tan ^2 x}=-\operatorname{cosec}^2 x
\end{aligned}
$

Example 4 

Find the derivative of
(i) $\frac{x^5-\cos x}{\sin x}$
(ii) $\frac{x+\cos x}{\tan x}$

Solution

(i) Let $h(x)=\frac{x^5-\cos x}{\sin x}$. We use the quotient rule on this function wherever it is defined.
$
h^{\prime}(x)=\frac{\left(x^5-\cos x\right)^{\prime} \sin x-\left(x^5-\cos x\right)(\sin x)^{\prime}}{(\sin x)^2}
$

$
\begin{aligned}
& =\frac{\left(5 x^4+\sin x\right) \sin x-\left(x^5-\cos x\right) \cos x}{\sin ^2 x} \\
& =\frac{-x^5 \cos x+5 x^4 \sin x+1}{(\sin x)^2}
\end{aligned}
$
(ii) We use quotient rule on the function $\frac{x+\cos x}{\tan x}$ wherever it is defined.
$
\begin{aligned}
h^{\prime}(x) & =\frac{(x+\cos x)^{\prime} \tan x-(x+\cos x)(\tan x)^{\prime}}{(\tan x)^2} \\
& =\frac{(1-\sin x) \tan x-(x+\cos x) \sec ^2 x}{(\tan x)^2}
\end{aligned}
$