In the previous section, we saw that derivative at a point, is the slope of the tangent at that point. In the section before that, we saw derivative as the rate of change. In this section, we will see an easy method to find the derivative at any given point.
It can be written in 11 steps:
1. In fig.13.30 below, the red curve is the graph of f(x).
 |
| Fig.13.30 |
• Two points P and Q are marked on the red curve.
♦ A green horizontal line is drawn through P.
♦ A green vertical line is drawn through Q.
♦ These horizontal and vertical lines meet at R
• Thus we get a right triangle PQR
2. A green dashed vertical line is drawn through P.
• This vertical line meets the x-axis at (a,0)
• So we can write:
The x-coordinate of P is ‘a’.
• If the x-coordinate is ‘a’, then obviously, the y-coordinate will be f(a)
• Thus we get the coordinates of P: (a,f(a))
3. Another green dashed vertical line is drawn through R.
• This vertical line meets the x-axis at ((a+h),0)
• So we can write:
The x-coordinate of Q is ‘(a+h)’.
• If the x-coordinate is ‘(a+h)’, then obviously, the y-coordinate will be f(a+h)
• Thus we get the coordinates of Q: ((a+h),f(a+h))
4. Using the x-coordinates of P and Q, we can find the horizontal distance between P and Q.
• We get:
Horizontal distance between P and Q = [(a+h) – a] = h
• That means, the length PR = h.
5. Using the y-coordinates of P and Q, we can find the vertical distance between P and Q.
• We get:
Vertical distance between P and Q = [f(a+h) – f(a)]
• That means, the length QR = [f(a+h) – f(a)].
6. Now we have the base and altitude of the right triangle PQR.
• So we can write the slope of the line PQ.
Slope of PQ = $\frac{[f(a+h) – f(a)]}{h}$
7. If Q is brought very close to P, then the slope calculated in (6) will be the slope of the tangent at P
• For bringing Q closer to P, we must decrease the length h. When h approaches zero, Q will be very close to P.
• h must become very close to zero. At the same time, it must not become exact zero. This can be written as:
$\lim_{h\rightarrow 0} h$
8. From (5), we have the length of the altitude:
[f(a+h) – f(a)]
• When h approaches zero, the base PR will become infinitesimal.
• There will be corresponding changes in the altitude also.
• When h approaches zero, the length of the altitude can be written as: $\lim_{h\rightarrow 0} {[f(a+h) – f(a)]}$
9. So now we can write the slope of the tangent at P:
$\frac{\lim_{h\rightarrow 0} {[f(a+h) – f(a)]}}{\lim_{h\rightarrow 0} h}$
• The limit is being applied to both numerator and denominator. So this can be written in a simplified form as:
Slope of tangent at P = $\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
10. But slope of the tangent at P is the derivative at P.
• Point P corresponds to (x = a)
• So we can write:
The result in (9) gives the derivative of f(x) at (x = a).
11. The derivative of f(x) at (x=a) is denoted as: f'(a).
• So we can write:
$$f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$$
Now we will see some solved examples:
Solved example 13.6
Find the derivative at x = 2 of the function f(x) = 3x.
Solution:
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 2 and f(x) = 3x.
• So we get:
$\begin{array}{ll}
{}&{f'(2)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(2+h) – f(2)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{3(2+h) – 3 × 2}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{6 + 3h – 6}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{3h}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {3 \lim_{h\rightarrow 0}{\left[\frac{h}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {3 \lim_{h\rightarrow 0}{1}}
&{} \\
{}&{}
& {~=~}& {3 × 1}
&{} \\
{}&{}
& {~=~}& {3}
&{} \\
\end{array}$
3. We see that the derivative is '3'.
• So we can write:
At the point where (x = 2), the rate of change of y with respect to x is 3.
• In fact, for this problem, the rate of change at every point will be 3.
4. We see that the derivative is '3'.
• So we can write:
At the point where (x = 2), the slope of the tangent is 3.
• In fact, for this problem, the slope of tangent at any point will be 3. The reason can be written in steps:
(i) For a straight line, the tangent at any point, will be the line itself.
(ii) Here, the slope of the line is 3. So the tangent at any point will also have the slope 3
Solved example 13.7
Find the derivative of the function f(x) = 2x2 + 3x - 5 at x = -1. Also prove that f'(0) + 3f'(-1) = 0
Solution:
Part (i):
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = -1 and f(x) = 2x2 + 3x - 5 .
• So we get:
$\begin{array}{ll}
{}&{f'(-1)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(-1+h) – f(-1)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2(-1+h)^2 + 3(-1+h) - 5] – [2×(-1)^2 + 3 × -1 - 5]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2(1 - 2h + h^2) + (-3+3h) - 5] – [2 - 3 - 5]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2 - 4h + 2h^2 -3+3h - 5 – 2 + 3 + 5}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{- h + 2h^2}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(2h - 1)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{[2h-1]}}
&{} \\
{}&{}
& {~=~}& {2 × 0~-~1}
&{} \\
{}&{}
& {~=~}& {-1}
&{} \\
\end{array}$
Part (ii):
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 0 and f(x) = 2x2 + 3x - 5.
• So we get:
$\begin{array}{ll}
{}&{f'(0)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(0+h) – f(0)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2(0+h)^2 + 3(0+h) - 5] – [2×(0)^2 + 3 × 0 ~- 5]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[2h^2 + 3h) - 5] – [2×0 + 3 × 0 ~- 5]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2h^2 + 3h - 5 + 5}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{2h^2 + 3h}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h(2h + 3)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[2h + 3 \right]}}
&{} \\
{}&{}
& {~=~}& {2 × 0 ~+ 3}
&{} \\
{}&{}
& {~=~}& {3}
&{} \\
\end{array}$
Part (iii):
1. From part (i), we have: f'(-1) = -1
2. From part (ii), we have: f'(0) = 3
3. So we get:
f'(0) + 3f'(-1) = [3 + (3 × -1)] = [3 - 3] = 0
At this stage, we are able to understand an important fact. It can be written in 2 steps:
(i) In the previous sections, we saw that:
Limits are subjected to various rules.
(ii) Those rules can be effectively used to evaluate the derivative.
Solved example 13.8
Find the derivative of sin x at x = 0
Solution:
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 0 and f(x) = sin x.
• So we get:
$\begin{array}{ll}
{}&{f'(0)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(0+h) – f(0)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\sin (0+h)] – [sin 0]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\sin h] – [sin 0]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[\sin h] – [0]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{\sin h}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {1}
&{} \\
\end{array}$
Solved example 13.9
Find the derivative of f(x) = 3 at x = 0 and at x = 3
Solution:
Part (i):
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 0 and f(x) = 3.
• So we get:
$\begin{array}{ll}
{}&{f'(0)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(0+h) – f(0)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[3] – [3]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{0}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {0}
&{} \\
\end{array}$
Part (ii):
1. We have: $f'(a)~=~\lim_{h\rightarrow 0}{\left[\frac{f(a+h) – f(a)}{h} \right]}$
2. In our present case, a = 3 and f(x) = 3.
• So we get:
$\begin{array}{ll}
{}&{f'(3)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(3+h) – f(3)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[3] – [3]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{0}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {0}
&{} \\
\end{array}$
◼ In this example, we see that:
♦ f’(0) = 0
♦ f’(3) = 0
• In fact, we can put any value for a. The derivative will be zero.
• The reason can be written in 2 steps:
1. f(x) = 3 is a constant function.
• That means, whatever be the value of x, the resulting f(x) will be the same.
• If there is no change in f(x), it means that, there is no rate of change.
♦ That means, rate of change is zero.
♦ That means, derivative is zero.
2. We can think in terms of slope of tangent also.
• For a straight line, the tangent at any point, will be the line itself.
• Here, the straight line is horizontal. So the tangent at any point will also be horizontal.
• For a horizontal line, the slope is zero.
• Now we know how to find the derivative at any given point.
• It would be convenient if we could obtain a general form.
• For example, we saw that:
Derivative of 2x2 + 3x -5 at (x = -1) is -1
• If we are asked to find the derivative of this function at say (x = 2), we will have to repeat all the steps using ‘2’.
• If we have a general form, we will be able to quickly find the derivative at any given point.
In the next section, we will see such a general form.
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