In the previous section,
we saw the general form of the derivative. We saw some solved examples also. In this section, we will see a few more solved examples. We will also see the graphs of derivatives.
Solved example 13.10
Find the derivative of f(x) = 10x.
Solution:
• In our present case, f(x) = 10x. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{10(x+h) – 10 x}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{10x + 10h – 10x}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{10h}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {10 \lim_{h\rightarrow 0}{\left[\frac{h}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {10 \lim_{h\rightarrow 0}{1}}
&{} \\
{}&{}
& {~=~}& {10 × 1}
&{} \\
{}&{}
& {~=~}& {10}
&{} \\
\end{array}$
◼ A graphical description can be written in 5 steps:
1. In the previous section, we saw that, f’(x) is a function. So we must be able to plot f’(x).
2. In fig.13.34 below, both f(x) and f’(x) are plotted in the same graph.
 |
| Fig.13.34 |
• In order to accommodate both functions in the same graph, the scale is changed to the following values:
♦ x-axis: 1 unit = 2
♦ y-axis: 1 unit = 10
3. The two graphs are plotted in different colors.
♦ The red line represents f(x)
♦ The green line represents f’(x)
4. f’(x) is a constant function. So, as expected, the green line is horizontal.
5. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at (x = 1.5). That is., we want f’(1.5)
• For that, we mark point A on the red line. Here, (x = 1.5)
• Next we draw a vertical line through A. This vertical line meets the green line at A'.
• f’(1.5) will be equal to the y-coordinate of A'.
(ii) Suppose that, we want the derivative of f(x) at (x = -2.3). That is., we want f’(-2.3)
• For that, we mark point B on the red line. Here, (x = -2.3)
• Next we draw a vertical line through B. This vertical line meets the green line at B'.
• f’(-2.3) will be equal to the y-coordinate of B'.
Solved example 13.11
Find the derivative of f(x) = x2 .
Solution:
• In our present case, f(x) = x2. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[(x+h)^2] – [x^2]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[(x^2 + 2hx + h^2)] – [x^2]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2hx + h^2}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(h +2x)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{[h + 2x]}}
&{} \\
{}&{}
& {~=~}& {2x}
&{} \\
\end{array}$
◼ A graphical description can be written in 5 steps:
1. In the previous section, we saw that, f’(x) is a function. So we must be able to plot f’(x).
2. In fig.13.35 below, both f(x) and f’(x) are plotted in the same graph.
 |
| Fig.13.35 |
3. The two graphs are plotted in different colors.
♦ The red curve represents f(x)
♦ The green line represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at (x = 3). That is., we want f’(3)
• For that, we mark point A on the red line. Here, (x = 3)
• Next we draw a vertical line through A. This vertical line meets the green line at A'.
• f’(3) will be equal to the y-coordinate of A'. In this case, it is 6.
•
We can verify this theoretically:
f`(3) = 2 × 3 = 6
(ii) Suppose that, we want the derivative of f(x) at (x = -2.5). That is., we want f’(-2.5)
• For that, we mark point B on the red line. Here, (x = -2.5)
• Next we draw a vertical line through B. This vertical line meets the green line at B'.
• f’(-2.5) will be equal to the y-coordinate of B'. In this case, it is -5.
•
We can verify this theoretically:
f`(-2.5) = 2 × -2.5 = -5
5. Further more, the reader may verify the tangents also:
•
Through the point A, draw a line at a slope of 6. This line will be tangential to f(x).
•
Through the point B, draw a line at a slope of -5. This line will be tangential to f(x).
Solved example 13.12
Find the derivative of the constant function f(x) = a for a fixed real number a.
Solution:
• In our present case, f(x) = a. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[a] – [a]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{0}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {0}
&{} \\
\end{array}$
Solved example 13.13
Find the derivative of f(x) = $\frac{1}{x}$.
Solution:
• In our present case, f(x) = $\frac{1}{x}$. So we get:
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\left[\frac{1}{x+h}\right] – \left[\frac{1}{x}\right]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\frac{x - x - h}{x(x+h)}}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-h}{xh(x+h)} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{-1}{x(x+h)} \right]}}
&{} \\
{}&{}
& {~=~}& {\frac{-1}{x(x+0)}}
&{} \\
{}&{}
& {~=~}& {-\frac{1}{x^2}}
&{} \\
\end{array}$
◼ A graphical description can be written in 5 steps:
1. In the previous section, we saw that, f’(x) is a function. So we must be able to plot f’(x).
2. In fig.13.36 below, both f(x) and f’(x) are plotted in the same graph.
 |
| Fig.13.36 |
3. The two graphs are plotted in different colors.
♦ The red curve represents f(x)
♦ The green curve represents f’(x)
4. Let us see the practical application of the graph. It can be written using two examples:
(i) Suppose that, we want the derivative of f(x) at (x = 1.5). That is., we want f’(1.5)
• For that, we mark point A on the red line. Here, (x = 1.5)
• Next we draw a vertical line through A. This vertical line meets the green line at A'.
• f’(1.5) will be equal to the y-coordinate of A'. In this case, it is -0.44.
•
We can verify this theoretically:
f`(1.5) = $-\frac{1}{1.5^2}$ = -0.44
(ii) Suppose that, we want the derivative of f(x) at (x = -2). That is., we want f’(-2)
• For that, we mark point B on the red line. Here, (x = -2)
• Next we draw a vertical line through B. This vertical line meets the green line at B'.
• f’(-2) will be equal to the y-coordinate of B'. In this case, it is -0.25.
•
We can verify this theoretically:
f`(-2) = $-\frac{1}{(-2)^2}$ = -0.25
5. Further more, the reader may verify the tangents also:
•
Through the point A, draw a line at a slope of 0.-44. This line will be tangential to f(x).
•
Through the point B, draw a line at a slope of -0.25. This line will be tangential to f(x).
In the next section, we will see algebra of derivative of functions.
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