21.15 - Derivatives of Exponential and Logarithmic Functions

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) = e^x, then it’s derivative is the same e^x.
• That is., if f(x) = e^x, then f'(x) = e^x.
    ♦ 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) = e^x.

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 = e^1.5 = 4.4816
• We will use a single decimal place and write:
e^1.5 = 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 e^x.
So f'(1.5) = e^1.5 = 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 e^1.5, which is the e^x at P(1.5,4.5).
(ii) We drew a line through P(1.5,4.5) at a slope equal to e^1.5.
(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 e^x.

Note that, the above demonstration is not a proof. We will see the actual proof in higher classes.

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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.

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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(e^x)    (iv) e^{cos 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, log_ee = 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.

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Link to a few more solved examples is given below:

Exercise 21.4

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In the next section, we will see logarithmic differentiation.

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