Question

If $y = {\left( {\dfrac{x}{n}} \right)^{nx}}\left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)$ then find the value of ${y'}(n)$ is given by

1. $\dfrac{1}{n}$
2. ${\left( {\dfrac{1}{n}} \right)^n}$
3. $\left( {\dfrac{{{n^2} + 1}}{n}} \right)$
4. ${\left( {\dfrac{1}{n}} \right)^n}\left( {\dfrac{{{n^2} + 1}}{n}} \right)$

Answer 1

Hint : Logs (or) logarithms are nothing but another way of expressing exponents. First, we will apply logs on both sides. Then we will use the logarithm rule of multiplication and power i.e. $\log (ab) = \log a + \log b$ and $\log {a^b} = b\log a$ . After this, we will apply the derivative of y with respect to x on both the sides. After evaluating this, substituting x = n in the given statement to get the new value of y. We also know the value of log1 = 0. Thus, using all this, we will get the final output.

Complete step-by-step answer :
Given that,
$y = {\left( {\dfrac{x}{n}} \right)^{nx}}\left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)$
Applying log on both the sides, we get,
$\Rightarrow \log y = \log \left( {{{\left( {\dfrac{x}{n}} \right)}^{nx}}\left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)} \right)$
We will use logarithm rule of multiplication:
$\log (ab) = \log a + \log b$
Applying this rule, we will get,
$\Rightarrow \log y = \log {\left( {\dfrac{x}{n}} \right)^{nx}} + \log \left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)$
We will also use logarithm rule of power:
$\log {a^b} = b\log a$
Applying this rule, we will get,
$\Rightarrow \log y = nx\log \left( {\dfrac{x}{n}} \right) + \log \left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)$
Applying derivative of y with respect to x on both the sides, we will get,
$\Rightarrow \dfrac{1}{y}{y'} = nx\left( {\dfrac{1}{{\dfrac{x}{n}}}} \right)\left( {\dfrac{1}{n}} \right) + \dfrac{1}{{1 + \log \left( {\dfrac{x}{n}} \right)}}\left( {\dfrac{n}{x}} \right)\left( {\dfrac{1}{n}} \right) + \log \left( {\dfrac{x}{n}} \right)(n)$
On evaluating this above equation, we will get,
$\Rightarrow \dfrac{1}{y}{y'} = nx\left( {\dfrac{n}{x}} \right)\left( {\dfrac{1}{n}} \right) + \dfrac{1}{{1 + \log \left( {\dfrac{x}{n}} \right)}}\left( {\dfrac{n}{x}} \right)\left( {\dfrac{1}{n}} \right) + \log \left( {\dfrac{x}{n}} \right)(n)$
$\Rightarrow \dfrac{1}{y}{y'} = n + \dfrac{1}{{1 + \log \left( {\dfrac{x}{n}} \right)}}\left( {\dfrac{1}{x}} \right) + \log \left( {\dfrac{x}{n}} \right)(n)$
Removing all extra brackets, we will get,
$\Rightarrow \dfrac{1}{y}{y'} = n + \dfrac{1}{{x\left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)}} + n\log \left( {\dfrac{x}{n}} \right)$
Taking term y from LHS to RHS, we will get,
\Rightarrow {y'}(x) = y\left\\{ {n + \dfrac{1}{{x\left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)}} + n\log \left( {\dfrac{x}{n}} \right)} \right\\} ------ (1)
Here, we are given,
$y = {\left( {\dfrac{x}{n}} \right)^{nx}}\left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)$
Let x=n
Substituting this value, we will get,
$\Rightarrow y = {\left( {\dfrac{n}{n}} \right)^{nx}}\left( {1 + \log \left( {\dfrac{n}{n}} \right)} \right)$
$\Rightarrow y = {1^{nn}}(1 + \log 1)$
We know that, log1=0 and so applying this, we will get,
$\Rightarrow y = 1(1 + 0)$
$\Rightarrow y = 1$
Thus, we have y=1 when x=n.
Now, using this value of y and x in equation (1), we will get,
\Rightarrow {y'}(x) = 1\left\\{ {n + \dfrac{1}{{n\left( {1 + \log \left( {\dfrac{n}{n}} \right)} \right)}} + n\log \left( {\dfrac{n}{n}} \right)} \right\\}
On evaluating this, we will get,
$\Rightarrow {y'}(n) = n + \dfrac{1}{{n(1 + \log 1)}} + n\log 1$
We know that, log1=0 and so applying this, we will get,
$\Rightarrow {y'}(n) = n + \dfrac{1}{{n(1 + 0)}} + n(0)$
$\Rightarrow {y'}(n) = n + \dfrac{1}{{n(1)}} + 0$
$\Rightarrow {y'}(n) = n + \dfrac{1}{n}$
$\Rightarrow {y'}(n) = \dfrac{{{n^2} + 1}}{n}$
Hence, for given $y = {\left( {\dfrac{x}{n}} \right)^{nx}}\left( {1 + \log \left( {\dfrac{x}{n}} \right)} \right)$ , the value of ${y'}(n) = \left( {\dfrac{{{n^2} + 1}}{n}} \right)$ .
So, the correct answer is “Option 4”.

Note : A logarithm is defined as the power to which number must be raised to get some other values. It is the most convenient way to express large numbers. There are three logarithm identities which one should know. They are:

1. Product rule: log (ab) = log a + log b
2. Quotient rule: $log\left( {\dfrac{a}{b}} \right) = loga-logb$
3. Power rule: $log({a^b}) = bloga$
   Logarithms are also said to be the inverse process of exponential. Also, log 0 is undefined. Because, we never get the value 0, by raising any value to the power of anything else.