Higher Secondary Mathematics: 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.

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: