21.21 Miscellaneous Examples

In the previous section, we completed a discussion on Rolle's theorem and mean value theorem. In this section, we will see some miscellaneous examples.

Solved example 21.66
Differentiate w.r.t x the following functions:
(i) $\rm{\sqrt{3x+2}~+~\frac{1}{\sqrt{2x^2 + 4}}}$
(ii) $\rm{e^{\sec^2 x}~+~3 \cos^{-1} x}$
(iii) $\rm{\log_7 (\log x)}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\sqrt{3x+2}~+~\frac{1}{\sqrt{2x^2 + 4}}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left(3x+2 \right)^{1/2} ~+~\left(2x^2 + 4 \right)^{-1/2}} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{1/2 \left(3x+2 \right)^{-1/2}.3 ~+~(-1/2)\left(2x^2 + 4 \right)^{-3/2}.4x} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{3}{2 \sqrt{3x+2}}~-~\frac{2x}{\left(\sqrt{2x^2 + 4} \right)^3}} \\
\end{array}$

Part (ii):
1. First we will split the given function
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{e^{\sec^2 x}~+~3 \cos^{-1} x} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{u+v} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{du}{dx}~+~\frac{dv}{dx}} \\
\end{array}$

2. Now we will find $\rm{\frac{du}{dx}}$:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{u} & {~=~} &{e^{\sec^2 x}} \\
{~\color{magenta} 2 } &{\implies} &{\log u} & {~=~} &{\sec^2 x.\log_e e} \\
{~\color{magenta} 3 } &{\implies} &{\log u} & {~=~} &{\sec^2 x} \\
{~\color{magenta} 4 } &{\implies} &{\frac{d}{dx} \left[\log u \right]} & {~=~} &{\frac{d}{dx} \left[\sec^2 x \right]} \\
{~\color{magenta} 5 } &{\implies} &{\frac{1}{u}.\frac{du}{dx}} & {~=~} &{2 \sec x . \sec x \tan x} \\
{~\color{magenta} 6 } &{\implies} &{\frac{du}{dx}} & {~=~} &{u. 2 \sec x . \sec x \tan x} \\
{~\color{magenta} 7 } &{{}} &{{}} & {~=~} &{2 e^{\sec^2 x} \sec^2 x \tan x} \\
\end{array}$

3. Now we will find $\rm{\frac{dv}{dx}}$:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{v} & {~=~} &{3 \cos^{-1} x} \\
{~\color{magenta} 2 } &{\implies} &{\frac{dv}{dx}} & {~=~} &{3 \frac{d}{dx} \left[\cos^{-1} x \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{3. \frac{(-1)}{\sqrt{1 – x^2}}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{-3}{\sqrt{1 – x^2}}} \\
\end{array}$

4. So based on (1), we get:
$\rm{\frac{dy}{dx}~=~\frac{du}{dx}~+~\frac{dv}{dx}}$

= $\rm{2 e^{\sec^2 x} \sec^2 x \tan x~-~\frac{3}{\sqrt{1 – x^2}}}$

Part (iii):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\log_7 (\log x)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{\log_e (\log x)}{\log_e 7}} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{1}{\log_e 7} \frac{d}{dx} \left[\log_e (\log x) \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{1}{\log_e 7} .\frac{1}{\log x}.\frac{1}{x}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{1}{x \log 7 \log x}} \\
\end{array}$


◼ Remarks:
• 2 (magenta color): Here we apply change of base rule.
• 5 (magenta color): When the base of logarithm is not specified, it implies that, the base is e.

Solved example 21.67
Differentiate w.r.t x the following functions:
(i) $\rm{\cos^{-1} (\sin x)}$
(ii) $\rm{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)}$
(iii) $\rm{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\cos^{-1} (\sin x)} \\
{~\color{magenta} 2 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{-1}{\sqrt{1 - \sin^2 x}}.\frac{d}{dx} (\sin x)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{-1}{\sqrt{\cos^2 x}}.\cos x} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{-1} \\
\end{array}$

Alternate method:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\cos^{-1} (\sin x)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\cos^{-1} \left[\cos\left(\frac{\pi}{2} ~-~x \right) \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{\pi}{2} ~-~x} \\
{~\color{magenta} 4 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{-1} \\
\end{array}$

Part (ii): $\rm{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)}$
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\tan^{-1} \left(\tan \frac{x}{2} \right)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{x}{2}} \\
{~\color{magenta} 4 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{1}{2}} \\
\end{array}$


◼ Remarks:
• 2 (magenta color): Here we use identities:
$\rm{\cos x ~=~ 1 - 2 \sin^2 \frac{x}{2}}$
$\rm{\cos x ~=~ 2 \cos^2 \frac{x}{2} - 1}$

Combining these two identities, we get:
$\rm{\tan \frac{x}{2}~=~\frac{\sin x}{1 + \cos x}}$

Part (iii): $\rm{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)}$

1. First we will simplify the expression.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Let}~2^x}    & {~=~}    &{t}    \\
{~\color{magenta}    2    }    &{{}}    &{\text{Given:}~\frac{2^{x+1}}{1+4^x}}    & {~=~}    &{\frac{2.2^x}{1+(2.2)^x}}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{2.2^x}{1+2^x .2^x}}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{2t}{1+t^2}}    \\
\end{array}$

2. Next we will write 't' in terms of 𝜃.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\text{Let}~t}    & {~=~}    &{\tan \frac{\theta}{2}}    \\
{~\color{magenta}    2    }    &{{}}    &{\text{Then:}~\frac{2t}{1+t^2}}    & {~=~}    &{\sin \theta}    \\
\end{array}$

(Identity 15. List of identities can be seen here) 

3. Next we will write 'y' in terms of 𝜃.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\sin^{-1} \left(\frac{2t}{1 + t^2} \right)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\sin^{-1}(\sin \theta)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\theta}    \\
\end{array}$

4. Next we will write 𝜃 in terms of 'x'.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{2^x}    & {~=~}    &{t}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\tan \frac{\theta}{2}}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{\theta}{2}}    & {~=~}    &{\tan^{-1}(2^x)}    \\
{~\color{magenta}    4    }    &{\implies}    &{\theta}    & {~=~}    &{2. \tan^{-1}(2^x)}    \\
\end{array}$

5. Based on (3), we can now connect 'y' and 'x'.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{y}    & {~=~}    &{2. tan^{-1}(2^x)}    \\
{~\color{magenta}    2    }    &{\implies}    &{\frac{dy}{dx}}    & {~=~}    &{2. \frac{1}{1 + (2^x)^2}.\frac{d}{dx}(2^x)}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{2. \frac{1}{1 + (2^2)^x}.\frac{d}{dx}(2^x)}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{\frac{2}{1 + 4^x}.\frac{d}{dx}(2^x)}    \\
\end{array}$

6. So our next task is to find the derivative of $\rm{2^x}$.
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{u}    & {~=~}    &{2^x}    \\
{~\color{magenta}    2    }    &{\implies}    &{\log u}    & {~=~}    &{x \log 2}    \\
{~\color{magenta}    3    }    &{\implies}    &{\frac{1}{u}.\frac{du}{dx}}    & {~=~}    &{1. \log 2}    \\
{~\color{magenta}    4    }    &{\implies}    &{\frac{du}{dx}}    & {~=~}    &{u.\log 2}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{2^x \log 2}    \\
\end{array}$

7. So from (5), we get:
$\begin{array}{ll} {~\color{magenta}    1    }    &{{}}    &{\frac{dy}{dx}}    & {~=~}    &{\frac{2}{1 + 4^x}.\frac{d}{dx}(2^x)}    \\
{~\color{magenta}    2    }    &{{}}    &{{}}    & {~=~}    &{\frac{2}{1 + 4^x}.(2^x) \log 2}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{\frac{2^{x+1} \log 2}{1 + 4^x}}    \\
\end{array}$

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In the next section, we will see a few more solved examples.

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21.13 - Properties of Logarithm

In the previous section, we saw some basic details about logarithmic functions. In this section, we will see some of their properties.

Property I: $\rm{\log_a p \,=\, \frac{\log_b p}{\log_b a}}$
   ♦ On the L.H.S, we have log of p to the base a.
   ♦ On the R.H.S, we have log of p to the base b.
• So this property gives us the method to connect two logs of the same number when the bases are different.
• This property is also known as change of base rule.
• We can write the proof in 5 steps:

1. Let $\rm{\log_a p = \alpha}$
Then $\rm{a^{\alpha} = p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b a = \gamma}$
Then $\rm{b^{\gamma} = a}$
4. Substituting (3) in (1), we get:
$\rm{\left(b^{\gamma} \right)^{\alpha} = p}$
⇒ $\rm{b^{\gamma \alpha} = p}$
5. Substituting (4) in (2), we get:
$\rm{b^{\beta} = p = b^{\gamma \alpha}}$
⇒ $\rm{\beta = \gamma \alpha}$
⇒ $\rm{\alpha = \frac{\beta}{\gamma}}$
⇒ $\rm{\log_a p = \frac{\log_b p}{\log_b a}}$

Property II: $\rm{\log_b pq \,=\, \log_b p + \log_b q}$
   ♦ On the L.H.S, we have log of a product pq.
   ♦ On the R.H.S, we have addition of two individual logs.
• So this property helps us to simplify logarithmic equations when products are involved.
• We can write the proof in 4 steps:
1. Let $\rm{\log_b pq = \alpha}$
Then $\rm{b^{\alpha} = pq}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b q = \gamma}$
Then $\rm{b^{\gamma} = q}$
4. Substituting (2) and (3) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\gamma}$
⇒ $\rm{b^\alpha = b^{\beta + \gamma}}$
⇒ $\rm{\alpha = \beta + \gamma}$
⇒ $\rm{\log_b pq = \log_b p + \log_b q}$

Property III: $\rm{\log_b \left(\frac{p}{q} \right) \,=\, \log_b p - \log_b q}$
   ♦ On the L.H.S, we have log of a ratio $\frac{p}{q}$.
   ♦ On the R.H.S, we have subtraction of one individual log from another.
• So this property helps us to simplify logarithmic equations when ratios are involved.
• We can write the proof in 4 steps:
1. Let $\rm{\log_b \left(\frac{p}{q} \right) = \alpha}$
Then $\rm{b^{\alpha} = \frac{p}{q}}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Let $\rm{\log_b q = \gamma}$
Then $\rm{b^{\gamma} = q}$
4. Substituting (2) and (3) in (1), we get:
$\rm{b^\alpha = \frac{b^\beta}{b^\gamma}}$
⇒ $\rm{b^\alpha = b^{\beta - \gamma}}$
⇒ $\rm{\alpha = \beta - \gamma}$
⇒ $\rm{\log_b pq = \log_b p - \log_b q}$

Property IV: $\rm{\log_b p^2 \,=\, 2 \log_b p}$
   ♦ On the L.H.S, we have log of a square.
   ♦ On the R.H.S, we have individual log multiplied by 2.
• So this property helps us to simplify logarithmic equations when squares are involved.
• We can write the proof in 3 steps:
1. Let $\rm{\log_b (p \times p) = \alpha}$
Then $\rm{b^{\alpha} = p \times p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Substituting (2) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\beta}$
⇒ $\rm{b^\alpha = b^{\beta + \beta}}$
⇒ $\rm{\alpha = \beta + \beta}$
⇒ $\rm{\log_b (p \times p) = \log_b p + \log_b p}$
⇒ $\rm{\log_b p^2 = 2 \log_b p}$

Property V: $\rm{\log_b p^3 \,=\, 3 \log_b p}$
   ♦ On the L.H.S, we have log of a cube.
   ♦ On the R.H.S, we have individual log multiplied by 3.
• So this property helps us to simplify logarithmic equations when cubes are involved.
• We can write the proof in 3 steps:
1. Let $\rm{\log_b (p \times p \times p) = \alpha}$
Then $\rm{b^{\alpha} = p \times p \times p}$
2. Let $\rm{\log_b p = \beta}$
Then $\rm{b^{\beta} = p}$
3. Substituting (2) in (1), we get:
$\rm{b^\alpha = b^\beta \times b^\beta \times b^\beta}$
⇒ $\rm{b^\alpha = b^{\beta + \beta + \beta}}$
⇒ $\rm{\alpha = \beta + \beta + \beta}$
⇒ $\rm{\log_b (p \times p \times p) = \log_b p + \log_b p + \log_b p}$
⇒ $\rm{\log_b p^3 = 3 \log_b p}$

Property VI: $\rm{\log_b p^n \,=\, n \log_b p}$
   ♦ On the L.H.S, we have log of a n^th power.
   ♦ On the R.H.S, we have individual log multiplied by n.
• So this property helps us to simplify logarithmic equations when n^th power is involved.
• The proof can be written using the principles of mathematical induction.

Property VII: $\rm{\log_p p \,=\, 1}$
   ♦ On the L.H.S, we have same number and base.
   ♦ On the R.H.S, we have 1.
We can write the proof as follows:
• Let $\rm{\log_p p = \alpha}$
• Then $\rm{p^{\alpha} = p}$
⇒ $\rm{p^{\alpha} = p^1}$
⇒ $\rm{\alpha = 1}$

Property VIII: $\rm{\log_b p \,=\, \frac{1}{\log_p b}}$
   ♦ On the L.H.S, we have base b and number p.
   ♦ On the R.H.S, we have base p and number b.
   ♦ So base and number are interchanged.
We can write the proof as follows:
• Applying the change of base rule we get:
$\rm{\log_b p \,=\, \frac{\log_p p}{\log_p b} \,=\, \frac{1}{\log_p b}}$ 

Property IX: $\rm{\log_p {p^x} \,=\, x}$
We can write the proof as follows:
• Let $\rm{\log_p p^x = \alpha}$
• Then $\rm{p^{\alpha} = p^x}$
⇒ $\rm{\alpha = x}$

• Note that, this property is used when we first begin to learn about logarithms. For example:
$\rm{\log_5 625 = \log_5 5^4 = 4}$

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Now we will see some solved examples

Solved example 21.43
Express as a single logarithm:
$\rm{2 \log x - 5 \log y + 3 \log z}$
Solution:
1. 2 log x = log x².
2. −5 log y = −log y⁵.
3. 3 log z = log z³.
4. log x² − log y⁵ + log z³ = $\rm{\frac{x^2 z^3}{y^5}}$

Solved example 21.44
Expand using the properties of logarithm:
$\rm{\log_5 \sqrt[3]{x}}$
Solution:
$\rm{\log_5 \sqrt[3]{x}}$
= $\rm{\log_5 \left(x^{\frac{1}{3}} \right)}$ 
= $\rm{\frac{1}{3} \log_5 x}$

Solved example 21.45
Evaluate log₅ 28
Solution:
1. We could find the solution easily if it is "25" instead of "28".
2. We cannot directly use the calculator because base is 5.
3. So we will apply the change of base rule. We get:
$\rm{\log_5 28 \,=\,\frac{\log_{10} 28}{\log_{10} 5}\,=\,\frac{1.44715}{0.69897}\,=\,2.07040}$

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In the next section, we will see exponential and logarithmic equations.

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