21.16 - Logarithmic Differentiation

In the previous section, we saw the methods to find the derivatives of exponential and logarithmic functions. In this section, we will see logarithmic differentiation.

The basic details can be written in 3 steps:
1. Consider the function: $\rm{y=f(x)=[u(x)]^{v(x)}}$
    ♦ The base is a function of x.
    ♦ Exponent is also a function of x.

2. For differentiating this type of functions, we take logarithm on both sides. We get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log(y)}    & {~=~}    &{\log \left([u(x)]^{v(x)} \right)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log(y)}    & {~=~}    &{v(x) \log [u(x)]}    \\
\end{array}$

3. Now we can apply the chain rule and obtain the derivative. This is shown below:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log(y)}    & {~=~}    &{v(x) \log [u(x)]}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{d}{dx} \left(\log(y) \right)}    & {~=~}    &{\frac{d}{dx} \left(v(x) \log [u(x)] \right)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{1}{y} . \frac{d}{dx}(y)}    & {~=~}    &{\frac{d}{dx} \left(v(x) \right) . \log[u(x)]\,+\,v(x) . \frac{d}{dx} \left(\log[u(x)] \right)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{v'(x) . \log[u(x)]\,+\,v(x) . \frac{1}{u(x)} \frac{d}{dx} \left(u(x) \right)}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{v'(x) . \log[u(x)]\,+\,v(x) . \frac{1}{u(x)} u'(x)}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{y \left[v'(x) . \log[u(x)]\,+\,v(x) . \frac{1}{u(x)} u'(x) \right]}    \\
\end{array}$                           

◼ Remarks:
• 2(magenta color):
    ♦ We are taking the logarithm of y. That means, we are taking the logarithm of f(x). So f(x) must be +ve.
    ♦ Similarly, we are taking the logarithm of u(x). So u(x) must be +ve.

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Let us see some solved examples:

Solved example 21.51
Find $\rm{\frac{dy}{dx}~\text{if}~y\,=\,\sqrt{\frac{(x-3)(x^2 + 4)}{3x^2 + 4x + 5}}}$
Solution:
1. Let us name the functions, so that, given y can be written in a short form:
    ♦ u(x) = x − 3
    ♦ v(x) = x²+4
    ♦ z(x) = 3x² + 4x + 5
• Then $\rm{y\,=\,\sqrt{\frac{u(x).v(x)}{z(x)}}}$ 

2. Taking logarithm on both sides, we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log(y)}    & {~=~}    &{\log \left(\sqrt{\frac{u(x).v(x)}{z(x)}} \right)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \log \left({\frac{u(x).v(x)}{z(x)}} \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \left(\log u(x)\,+\,\log v(x) \,-\, \log z(x) \right)}    \\
\end{array}$                           

3. Now we can find the derivative as follows:

$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log(y)}    & {~=~}    &{\frac{1}{2} \left(\log u(x)\,+\,\log v(x) \,-\, \log z(x) \right)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{d}{dx} \left(\log(y) \right)}    & {~=~}    &{\frac{d}{dx} \left[\frac{1}{2} \left(\log u(x)\,+\,\log v(x) \,-\, \log z(x) \right) \right]}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{1}{y}.\frac{d}{dx} (y) }    & {~=~}    &{\frac{1}{2} \frac{d}{dx} \left[\log u(x)\,+\,\log v(x) \,-\, \log z(x) \right]}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{\frac{1}{2}  \left[\frac{d}{dx} \log u(x)\,+\,\frac{d}{dx} \log v(x) \,-\, \frac{d}{dx} \log z(x) \right]}    \\
{~\color{magenta}    5    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{\frac{1}{2}  \left[\frac{1}{u(x)} . u'(x)\,+\,\frac{1}{v(x)} . v'(x) \,-\, \frac{1}{z(x)} . z'(x) \right]}    \\
{~\color{magenta}    6    }    &{\implies}    &{\frac{1}{y} . \frac{dy}{dx}}    & {~=~}    &{\frac{1}{2}  \left[\frac{1}{x-3} . (1)\,+\,\frac{1}{x^2 - 4} . (2x) \,-\, \frac{1}{3x^2 + 4x + 5} . (6x+4) \right]}    \\
{~\color{magenta}    7    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{y}{2} \left[\frac{1}{x-3} \,+\,\frac{2x}{x^2 - 4} \,-\, \frac{ 6x+4}{3x^2 + 4x + 5} \right]}    \\
{~\color{magenta}    8    }    &{{}}    &{{}}    & {~=~}    &{\frac{1}{2} \sqrt{\frac{(x-3)(x^2 + 4)}{3x^2 + 4x + 5}} \left[\frac{1}{x-3} \,+\,\frac{2x}{x^2 - 4} \,-\, \frac{ 6x+4}{3x^2 + 4x + 5} \right]}    \\
\end{array}$

Solved example 21.52
Differentiate a^x w.r.t. x, where a is a positive constant.
Solution:
1. Let y = a^x.
2. Taking logarithm on both sides, we get:
log y = x log a.
3. Now we can find the derivative as follows:

Alternate method:
• First we will prove the identity: u = e^{log u}.
• Proof can be written in (ii) steps:
(i) Let v = e^{log u}.
(ii) Taking log on both sides, we get:
log v = log(e^{log u}) = log u.log e= log u.(1) = log u
⇒ log v = log u
⇒ v = u
⇒ e^{log u} = u

1. Let y = a^x.
2. $\rm{a^x~\text{is same as}~e^{\log a^x}}$
3. So we get: $\rm{y = e^{\log a^x}}$
4. Now we can find the derivative as follows:

 

Solved example 21.53
Differentiate x^{sin x} , x>0 w.r.t. x.
Solution:
1. Let y = x^{sin x}, x>0
2. Taking logarithm on both sides, we get:
log y = sin x log x.
3. Now we can find the derivative as follows:

Solved example 21.54
Find $\rm{\frac{dy}{dx}~\text{if}~y^x\,+\,x^y\,+\,x^x\,=\,a^b}$.
Solution:
1. Given that: y^x + x^y + x^x = a^b.
2. Let us write the L.H.S as the sum of three functions:
u(x) + v(x) + w(x) = a^b.
• Where:
   ♦ u(x) = y^x .
   ♦ v(x) = x^y .
   ♦ w(x) = x^x .
3. Now we get:

4. First we will find $\rm{\frac{du}{dx}}$:

5. Next we will find $\rm{\frac{dv}{dx}}$:

6. Finally we will find $\rm{\frac{dw}{dx}}$:

7. Substituting the above results in (3), we get:

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The link below gives a few more solved examples:

Exercise 21.5

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In the next section, we will see derivatives of functions in parametric forms.

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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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21.14 - Solution of Exponential and Logarithmic Functions

In the previous section, we saw properties of logarithms. In this section, we will see some solved examples which demonstrate the process of solving exponential and logarithmic equations.

Solved example 21.46
Solve the equation: log(6x) − log(4-x) = log 3
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log (6x) - \log (4-x)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log \left(\frac{6x}{4-x} \right)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{6x}{4-x}}    & {~=~}    &{3}    \\
{~\color{magenta}    4    }    &{\implies}    &{6x}    & {~=~}    &{12 – 3x}    \\
{~\color{magenta}    5    }    &{\implies}    &{9x}    & {~=~}    &{12}    \\
{~\color{magenta}    6    }    &{\implies}    &{x}    & {~=~}    &{\frac{4}{3}}    \\
\end{array}$                           

Check:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\log (6(4/3)) - \log (4- 4/3)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log (8) - \log (8/3)}    & {~=~}    &{\log (3)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{8}{8/3}}    & {~=~}    &{3}    \\
{~\color{magenta}    4    }    &{\implies}    &{3}    & {~=~}    &{3}    \\
\end{array}$

Solved example 21.47
Solve the equation: ln(4 −3x) − ln(7x) = ln(11)
Solution:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\ln(4-3x) - \ln(7x)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\ln\left[\frac{4-3x}{7x} \right]}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{4-3x}{7x}}    & {~=~}    &{11}    \\
{~\color{magenta}    4    }    &{\implies}    &{4 – 3x}    & {~=~}    &{77x}    \\
{~\color{magenta}    5    }    &{\implies}    &{80x}    & {~=~}    &{4}    \\
{~\color{magenta}    6    }    &{\implies}    &{x}    & {~=~}    &{\frac{1}{20}}    \\
\end{array}$                            
 

Check:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\ln(4-3x) - \ln(7x)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\ln(4-3(1/20)) - \ln(7(1/20))}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    3    }    &{\implies}    &{\ln(77/20) - \ln(7/20)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\ln(77) - \ln(20) - \ln(7) + \ln(20)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    5    }    &{\implies}    &{\ln(77) - \ln(7)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    6    }    &{\implies}    &{\ln(77/7)}    & {~=~}    &{\ln(11)}    \\
{~\color{magenta}    7    }    &{\implies}    &{\ln(11)}    & {~=~}    &{\ln(11)}    \\
\end{array}$

Solved example 21.48
Solve the equation: log₈ (4x + 1) = −1
Solution:

Check:

Solved example 21.49
Solve the equation: 2e^3y+8 − 11e^5−10y = 0
Solution:

Check:

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We have seen the process of solving exponential and logarithmic equations. The reader is advised to try a large number of practice problems in this category. In the next section, we will see derivatives of exponential and logarithmic functions.

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