In the previous section,
we completed a discussion on exponential and
logarithmic functions. In this section, we will see the derivatives of those two functions.
First we will see the derivative of exponential function.
• If f(x) = ex, then it’s derivative is the same ex.
• That is., if f(x) = ex, then f'(x) = ex.
♦ In other words, $\rm{\text{if}~y = e^x,~\text{then},~\frac{dy}{dx} = e^x}$
• We will see the proof in higher classes. At present, we will see a simple application of this derivative. It can be written in 5 steps:
1. The red curve in fig.21.20 below shows the graph of f(x) = ex.
 |
| Fig.21.20 |
2. Mark any convenient point on the curve. Let us mark the point with x-coordinate 1.5.
• Since the x-coordinate is 1.5,
y-coordinate = e1.5 = 4.4816
• We will use a single decimal place and write:
e1.5 = 4.5.
• So the coordinates are (1.5,4.5). We will name this point as P
3. Next, we want the derivative of f(x) at x = 1.5
• That is, we want f'(1.5).
• We wrote that, f'(x) is the same ex.
So f'(1.5) = e1.5 = 4.5
4. We know that, f'(1.5) will be the slope of the tangent at x = 1.5
• Let us draw a line through P, at a slope of f'(1.5).
We have a point P(1.5,4.5) and slope 4.5. So the equation of this line will be:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y-y_1} & {~=~} &{\text{slope} \times (x - x_1)} \\
{~\color{magenta} 2 } &{\implies} &{y-4.5} & {~=~} &{4.5 \times (x – 1.5)} \\
{~\color{magenta} 3 } &{\implies} &{y-4.5} & {~=~} &{4.5 x \,–\, 6.75} \\
{~\color{magenta} 4 } &{\implies} &{y} & {~=~} &{4.5 x \,–\, 2.25} \\
\end{array}$
• Let us plot this line. It is shown in green color in fig.21.20 above.
(note that, the y-intercept of the green line in the fig.21.20 is −2.25)
• We see that, the green line is the tangent at P.
5. Let us write a summary:
(i) We calculated e1.5, which is the ex at P(1.5,4.5).
(ii) We drew a line through P(1.5,4.5) at a slope equal to e1.5.
(iii) That line happens to be the tangent at P. So e 1.5 is the derivative at P
(iv) Therefore, the general form of the derivative is ex.
Note that, the above demonstration is not a proof. We will see the actual proof in higher classes.
Now we will see the derivative of log function.
• If f(x) = log x, then it’s derivative is $\rm{\frac{1}{x}}$.
• That is., if f(x) = log x, then f'(x) = $\rm{\mathbf{{\frac{1}{x}}}}$.
♦ In other words, $\rm{\text{if}~y = \log x,~\text{then},~\frac{dy}{dx} = \frac{1}{x}}$
(Recall that, in this chapter, when we write log x, it means, base is e)
•
We will see the proof in higher classes. At present, we will see a
simple application of this derivative. It can be written in 5 steps:
1. The red curve in fig.21.21 below shows the graph of f(x) = log x.
 |
| Fig.21.21 |
2. Mark any convenient point on the curve. Let us mark the point with x-coordinate 2.0
• Since the x-coordinate is 2.0,
y-coordinate = log 2 = 0.6931
• We will use a single decimal place and write:
log = 0.7
• So the coordinates are (2,0.7). We will name this point as P
3. Next, we want the derivative of f(x) at x = 2
• That is, we want f'(2).
• We wrote that, f'(x) is $\frac{1}{x}$.
So f'(2) = $\rm{\frac{1}{2}}$ = 0.5
4. We know that, f'(2) will be the slope of the tangent at x = 2
• Let us draw a line through P, at a slope of f'(2).
We have a point P(2,0.7) and slope 0.5. So the equation of this line will be:
$\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{y-y_1} &
{~=~} &{\text{slope} \times (x - x_1)} \\
{~\color{magenta} 2 } &{\implies} &{y-0.7} & {~=~} &{0.5 \times (x – 2)} \\
{~\color{magenta} 3 } &{\implies} &{y-0.7} & {~=~} &{0.5 x \,–\, 1} \\
{~\color{magenta} 4 } &{\implies} &{y} & {~=~} &{0.5 x \,–\, 0.3} \\
\end{array}$
• Let us plot this line. It is shown in green color in fig.21.21 above.
(note that, the y-intercept of the green line in the fig.21.20 is −0.3)
• We see that, the green line is the tangent at P.
5. Let us write a summary:
(i) We calculated $\frac{1}{x}$ at P(2,0.7).
(ii) We drew a line through P(2,0.7) at a slope equal to $\frac{1}{2}$.
(iii) That line happens to be the tangent at P. So $\frac{1}{x}$ is the derivative at P
(iv) Therefore, the general form of the derivative is $\frac{1}{x}$.
Note that, the above demonstration is not a proof. We will see the actual proof in higher classes.
Now we will see some solved examples:
Solved example 21.50
Differentiate the following w.r.t to x:
(i) e−x (ii) sin(log x), x>0 (iii) cos−1(ex) (iv) ecos x.
Solution:
Part (i):
◼ Remarks:
• 3(Magenta color): Here we apply chain rule.
Alternate method:
◼ Remarks:
• 2(Magenta color): Here we take logarithm on both sides.
• 4(Magenta color): Here we apply chain rule.
• 5(Magenta color): Here we apply the fact that, logee = 1.
Part (ii):
◼ Remarks:
• 3(Magenta color): Here we apply chain rule.
Part (iii):
◼ Remarks:
• 3(Magenta color): Here we apply chain rule.
Part (iv):
◼ Remarks:
• 3(Magenta color): Here we apply chain rule.
Link to a few more solved examples is given below:
Exercise 21.4
In the next section, we will see logarithmic differentiation.
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