Chapter 13.18 - Miscellaneous Examples

In the previous section, we completed a discussion on the derivatives of trigonometric functions. In this section, we will see some miscellaneous examples.

Solved example 13.19
Find the derivative of f from the first principle, where f is given by
(i) $f(x) = \frac{2x+3}{x-2}$    (ii) $f(x) = x + \frac{1}{x}$
Solution:
Part (i):
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2(x+h) + 3}{(x+h) - 2}~–~ \frac{2x+3}{x-2}}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2x + 2h + 3}{x+h - 2}~–~ \frac{2x+3}{x-2}}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{(2x + 2h + 3)(x-2) ~-~(x+h-2)(2x+3)}{(x+h - 2)(x-2)}}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{2x^2 - 4x + 2hx - 4h +3x - 6 - 2x^2 - 2hx + 4x - 3x - 3h +6}{(x+h - 2)(x-2)}}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{  - 4h - 3h }{(x+h - 2)(x-2)}}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-7h}{h(x+h - 2)(x-2)} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-7}{(x+h - 2)(x-2)} \right]}} &{} \\ {}&{} & {~=~}& {\frac{-7}{(x - 2)(x-2)}} &{} \\ {}&{} & {~=~}& {\frac{-7}{(x - 2)^2}} &{} \\ \end{array}$

Part (ii):
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[(x+h) + \frac{1}{(x+h)} \right]~-~\left[x + \frac{1}{x} \right]}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\frac{(x+h)^2 + 1}{(x+h)} \right]~-~\left[\frac{x^2 + 1}{x} \right]}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{\left[(x+h)^2 + 1 \right]x~-~\left[(x^2 + 1)(x+h) \right]}{(x+h)x}}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[(x+h)^2 + 1 \right]x~-~\left[(x^2 + 1)(x+h) \right]}{(x+h)xh} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[x(x+h)^2 + x \right]~-~\left[x^3 + x^2 h + x + h \right]}{(x+h)xh} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[x^3 + 2 x^2 h + h^2 x + x \right]~-~\left[x^3 + x^2 h + x + h \right]}{(x+h)xh} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^3 + 2 x^2 h + h^2 x + x ~-~x^3 - x^2 h - x - h }{(x+h)xh} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^2 h + h^2 x - h }{(x+h)xh} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(x^2  + h x - 1) }{(x+h)xh} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x^2  + h x - 1 }{(x+h)x} \right]}} &{} \\ {}&{} & {~=~}& {\frac{x^2  + (0) x - 1 }{(x+0)x}} &{} \\ {}&{} & {~=~}& {\frac{x^2 - 1 }{x^2}} &{} \\ {}&{} & {~=~}& {1 - \frac{1 }{x^2}} &{} \\ \end{array}$

Solved example 13.20
Find the derivative of f(x) from the first principle, where f(x) is
(i) sin x + cos x     (ii) x sin x
Solution:
Part (i):
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\sin (x+h) + \cos (x+h) \right]~-~\left[\sin x + \cos x \right]}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h) + \cos (x+h) - \sin x - \cos x}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin (x+h) + \cos (x+h) - \sin x - \cos x}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\sin (x+h) - \sin x \right]+ \left[\cos (x+h) - \cos x \right]}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[2 \cos \frac{2x+h}{2} \sin \frac{h}{2} \right]+ \left[-2 \sin \frac{2x+h}{2} \sin \frac{h}{2} \right]}{h} \right]}~\color{green}{\text{- - - I}}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2 \sin \frac{h}{2} \left[\cos \frac{2x+h}{2}~-~\sin \frac{2x+h}{2} \right]}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin \frac{h}{2}}{h/2} \right]}~ × ~\lim_{h\rightarrow 0}{\left[\cos \frac{2x+h}{2}~-~\sin \frac{2x+h}{2} \right]}} &{} \\ {}&{} & {~=~}& {1~ × ~\left[\cos \frac{2x+0}{2}~-~\sin \frac{2x+0}{2} \right]} &{} \\ {}&{} & {~=~}& {1~ × ~\left[\cos x~-~\sin x \right]} &{} \\ {}&{} & {~=~}& {\cos x~-~\sin x} &{} \\ \end{array}$

◼ Remarks:
• Line marked as I:
Here we use the following identities:
    ♦ sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2
    ♦ cos A - cos B = -2 sin (A+B)/2 sin (A-B)/2

Part (ii):
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(x+h) \sin (x+h) ~-~x \sin x}{h}   \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x \sin (x+h) + h \sin (x+h) - x \sin x}{h}\right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x\left[\sin (x+h)  -  \sin x \right] + h \sin (x+h)}{h}   \right]}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x\left[2 \cos \frac{2x+h}{2} \sin \frac{h}{2} \right] + h \sin (x+h)}{h} \right]}~\color{green}{\text{- - - I}}} &{} \\ {}&{} & {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2x \cos \frac{2x+h}{2} \sin \frac{h}{2}}{h} \right]}~+~\lim_{h\rightarrow 0}{\left[\frac{h \sin (x+h)}{h} \right]}} &{} \\ {}&{} & {~=~}& {\left[\lim_{h\rightarrow 0}{\left[x \cos \frac{2x+h}{2} \right]}~ × ~\lim_{h\rightarrow 0}{\left[\frac{\sin \frac{h}{2}}{h/2} \right]} \right] ~+~\lim_{h\rightarrow 0}{\left[ \sin (x+h) \right]}} &{} \\ {}&{} & {~=~}& {\left[{\left[x \cos \frac{2x+0}{2} \right]}~ × ~1 \right] ~+~{\left[\sin (x+0) \right]}} &{} \\ {}&{} & {~=~}& {x \cos x + \sin x} &{} \\ \end{array}$

◼ Remarks:
• Line marked as I:
Here we use the following identity:
    ♦ sin A - sin B = 2 cos (A+B)/2 sin (A-B)/2

Solved example 13.21
Compute the derivative of
(i) f(x) = sin 2x  (ii) g(x) = cot x
Solution:
Part (i):
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {(\sin 2x)'} &{} \\ {}&{} & {~=~}& {(2 \sin x \cos x)'} &{} \\ {}&{} & {~=~}& {(2 \sin x)' (\cos x)~+~(2\sin x)(\cos x)' \color{green}{\text{- - - I}}} &{} \\ {}&{} & {~=~}& {(2 \sin x)' (\cos x)~+~(2\sin x)(-\sin x)} &{} \\ {}&{} & {~=~}& {(2 \sin x)' (\cos x)~-~2 \sin^2 x} &{} \\ {}&{} & {~=~}& {\left[(2)' \sin x ~+~ (2) (\sin x)' \right]\cos x~-~2 \sin^2 x} &{} \\ {}&{} & {~=~}& {\left[0 × \sin x ~+~ (2) (\cos x) \right]\cos x~-~2 \sin^2 x} &{} \\ {}&{} & {~=~}& {\left[2 \cos x \right]\cos x~-~2 \sin^2 x} &{} \\ {}&{} & {~=~}& {2 \cos^2 x~-~2 \sin^2 x} &{} \\ {}&{} & {~=~}& {2 (\cos^2 x - \sin^2 x)} &{} \\ \end{array}$

◼ Remarks:
• Line marked as I:
Here we use the product rule: (uv)' = u'v + uv'

Part (ii):
• Here we want to find the derivative of cot x.
• This is already done before.
• See question number 11(v) of Exercise 13.2.

Solved example 13.22
Find the derivative of
(i) $\frac{x^5 - \cos x}{\sin x}$  (ii) $\frac{x + \cos x}{\tan x}$
Solution:
Part (i):
1. Here we can use quotient rule:
$\left(\frac{u}{v}\right)' = \frac{u' v - u v'}{u^2}$
• In our present case,
   ♦ u = x⁵ - cos x
   ♦ v = sin x
2. Thus we get:
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {\frac{\left[(x^5 - \cos x)' \sin x \right] ~-~\left[(x^5 - \cos x) (\sin x)' \right]}{\sin^2 x}} &{} \\ {}&{} & {~=~}& {\frac{\left[(5 x^4 + \sin x) \sin x \right] ~-~\left[(x^5 - \cos x) (\cos x) \right]}{\sin^2 x}} &{} \\ {}&{} & {~=~}& {\frac{\left[5 x^4 \sin x + \sin^2 x \right] ~-~\left[x^5 \cos x - \cos^2 x \right]}{\sin^2 x}} &{} \\ {}&{} & {~=~}& {\frac{5 x^4 \sin x + \sin^2 x  - x^5 \cos x + \cos^2 x}{\sin^2 x}} &{} \\ {}&{} & {~=~}& {\frac{5 x^4 \sin x - x^5 \cos x + 1}{\sin^2 x}} &{} \\ {}&{} & {~=~}& {\frac{- x^5 \cos x + 5 x^4 \sin x + 1}{\sin^2 x}} &{} \\ \end{array}$

Part (ii):
1. Here we can use quotient rule:
$\left(\frac{u}{v}\right)' = \frac{u' v - u v'}{u^2}$
• In our present case,
   ♦ u = x + cos x
   ♦ v = tan x
2. Thus we get:
$\begin{array}{ll} {}&{f'(x)} & {~=~}& {\frac{\left[(x + \cos x)' \tan x \right] ~-~\left[(x + \cos x) (\tan x)' \right]}{\tan^2 x}} &{} \\ {}&{} & {~=~}& {\frac{\left[(1- \sin x) \tan x \right] ~-~\left[(x + \cos x) (\sec^2 x) \right]}{\tan^2 x}} &{} \\ \end{array}$

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Link to a few more miscellaneous examples is given below:

Miscellaneous Exercise on chapter 13

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In the next chapter, we will see mathematical reasoning.

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