In the previous section, we saw the graphical representation of the derivative. We saw some solved examples also. In this section, we will see algebra of derivatives.
•
We obtained derivative by using the principles of limits.
So we can expect:
♦ algebra of derivatives
♦ to be similar to
♦ algebra of limits.
•
For example, let us consider the sum of two functions. We know that:
♦ limit of the sum [f(x) + g(x)]
[This sum is written as: (f+g)(x)]
♦ is equal to
♦ limit of f(x) + limit of g(x)
•
We can extend this to derivatives:
♦ derivative of the sum (f+g)(x)
♦ is equal to
♦ derivative of f(x) + derivative of g(x)
•
That is, (f+g)'(x) = f'(x) + g'(x)
•
The proof can be written in 10 steps:
1. Suppose that, x1 is an allowable input x-value for the two functions f and g
•
First, calculate f(x1)
•
Next, calculate g(x1)
•
Next, calculate the sum [f(x1) + g(x1)]
•
Finally, calculate (f+g)(x1). It will be same as the above sum.
2. This can be illustrated using an example:
♦ Let f(x) = x2 + 1
♦ Let g(x) = 2x + 3
•
Then (f+g)(x) = x2 + 2x + 4
3. Let us try an input x-value of ‘2’
♦ f(2) = 22 + 1 = 5
♦ g(2) = 2 × 2 + 3 = 7
♦ f(2) + g(2) = 5 + 7 = 12
4. Let us try the same input x-value for (f+g)(x). We get:
(f+g)(2) = (22 + 2 × 2 + 4) = (4 + 4 + 4) = 12
5. By comparing the results in (3) and (4), we see that:
The statement in (1) is true.
6. Now we revisit the familiar fig.13.31 in section 13.13. Recall that, we used that fig.13.31 to obtain the derivative. We are going to use a modified version of that fig.13.31. It is shown below:
 |
| Fig.13.37 |
7. This time, the red curve represents the graph of (f+g)x.
♦ So the coordinates of P will be: (x,(f+g)(x))
♦ Also, the coordinates of Q will be: ((x+h), (f+g)(x+h))
8. Using the coordinates, we get:
♦ Altitude QR = [(f+g)(x+h) – (f+g)(x)]
♦ Base PR = [(x+h) – x] = h
9. So the slope of the yellow line = $\frac{(f+g)(x+h) – (f+g)(x)}{h}$
10. So slope of the tangent at P will be:
(f+g)'(x) =$\lim_{h\rightarrow 0}{\left[\frac{(f+g)(x+h) – (f+g)(x)}{h} \right]}$
•
This can be simplified as follows:
$\begin{array}{ll}
{}&{(f+g)'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{(f+g)(x+h) – (f+g)(x)}{h} \right]}}
&{} \\
{}&{}
&
{~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[f(x+h) + g(x+h)] – [f(x)
+ g(x)]}{h} \right]~\color{green}{\text{- - - I}}}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[f(x+h) - f(x) ] + [g(x+h) - g(x)]}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[f(x+h) - f(x) ]}{h}~+~\frac{[g(x+h) - g(x)]}{h} \right]}}
&{} \\
{}&{}
&
{~=~}& {\lim_{h\rightarrow 0}{\left[\frac{[f(x+h) - f(x)
]}{h}\right]}~+~\lim_{h\rightarrow 0}{\left[\frac{[g(x+h) - g(x)]}{h}
\right]}}
&{} \\
{}&{}
& {~=~}& {f'(x)~+~g'(x)}
&{} \\
\end{array}$
◼ Remarks:
•
Line marked as I:
In this line, we apply the result in (5).
• Using the same technique, we will be able to prove that:
(f-g)'(x) = f'(x) - g'(x)
•
For multiplication, we use the formula:
(f.g)'(x) = f'(x).g(x) + f(x).g'(x)
•
We will see the proof in higher classes.
•
This formula is easier to remember, if we use the following substitutions:
♦ f(x) = u and f'(x) = u'
♦ g(x) = v and g'(x) = v'
♦ (f.g)'(x) = (uv)'
•
Now the formula becomes: (uv)' = u'v + uv'
•
This is known as Leibnitz rule for differentiating product of functions.
• For division, we use the formula:
$\left(\frac{f}{g} \right)'(x)~=~ \frac{f'(x).g(x) - f(x).g'(x)}{\left(g(x) \right)^2}$
•
This formula is easier to remember, if we use the same substitutions as for multiplication. Then the formula becomes:
$\left(\frac{u}{v} \right)'(x)~=~ \frac{u'.v - u.v'}{v^2}$
•
Now we have the formulas for addition, subtraction, multiplication and division.
•
The four formulas are together known as theorem 5.
Using theorem 5, we can compute some common derivatives.
◼ First we will find f’(x) when f(x) = x
$\begin{array}{ll}
{}&{f'(x)}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{f(x+h) – f(x)}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{x+h - x}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[\frac{h}{h} \right]}}
&{} \\
{}&{}
& {~=~}& {\lim_{h\rightarrow 0}{\left[1 \right]}}
&{} \\
{}&{}
& {~=~}& {1}
&{} \\
\end{array}$
Now we will calculate f'(x) when f(x) = 10x
$\begin{array}{ll}
{}&{f'(10x)}
& {~=~}& {f'(x) + f'(x) + f'(x)~ + ~.~.~.~+f'(x)~\text{(10 terms)}~\color{green}{\text{- - - I}}}
&{} \\
{}&{}
& {~=~}& {1 + 1 + 1~+ ~.~.~.~+~1~\text{(10 terms)}}
&{} \\
{}&{}
& {~=~}& {10}
&{} \\
\end{array}$
◼ Remarks:
•
Line marked as I:
In this line, we apply the sum rule.
• The same result can be obtained by applying product rule also. For that, we write:
♦ f(x) = u = 10
♦ g(x) = v = x
♦ f(x).g(x) = uv = 10x
•
Applying the product rule, we get:
$\begin{array}{ll}
{}&{(uv)'}
& {~=~}& {u' v + u v'}
&{} \\
{}&{}
& {~=~}& {f'(10) × x + 10 × g'(x)}
&{} \\
{}&{}
& {~=~}& {0 × x + 10 × 1~\color{green}{\text{- - - I}}}
&{} \\
{}&{}
& {~=~}& {10}
&{} \\
\end{array}$
◼ Remarks:
•
Line marked as I:
In this line, we apply the following facts:
♦ Derivative of a constant function is zero.
♦ Derivative of x is 1.
• In the same way, we can obtain the derivative of f(x) = x2. For that, we write:
♦ f(x) = u = x
♦ g(x) = v = x
♦ f(x).g(x) = uv = x2
•
Applying the product rule, we get:
$\begin{array}{ll}
{}&{(uv)'}
& {~=~}& {u' v + u v'}
&{} \\
{}&{}
& {~=~}& {f'(x) × x + x × g'(x)}
&{} \\
{}&{}
& {~=~}& {1 × x + x × 1~\color{green}{\text{- - - I}}}
&{} \\
{}&{}
& {~=~}& {x + x}
&{} \\
{}&{}
& {~=~}& {2x}
&{} \\
\end{array}$
• Line marked as I:
In this line, we apply the following fact:
♦ Derivative of x is 1.
Now we know the derivative f'(x) when f(x) = x2. In the next section, we will see the derivative when the power of x is any +ve integer.
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