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:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\frac{\left[(x + \cos x)' \tan x \right] ~-~\left[(x + \cos x) (\tan x)' \right]}{\tan^2 x}} &{} \\
{}&{}
& {~=~}& {\frac{\left[(1- \sin x) \tan x \right] ~-~\left[(x + \cos x) (\sec^2 x) \right]}{\tan^2 x}} &{} \\
\end{array}$
Link to a few more miscellaneous examples is given below:
Miscellaneous Exercise on chapter 13
In the next chapter, we will see mathematical reasoning.
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