Question

If \[y=1+\dfrac{1}{x}+\frac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}}+.....\infty \] with \[\left| x \right|>1\], then \[\dfrac{dy}{dx}\] is
A. \[\dfrac{{{x}^{2}}}{{{y}^{2}}}\]
B. \[{{x}^{2}}{{y}^{2}}\]
C. \[\dfrac{{{y}^{2}}}{{{x}^{2}}}\]
D. \[-\dfrac{{{y}^{2}}}{{{x}^{2}}}\]

Answer 1

Hint: Firstly identify the type of series given in the question. After that use the sum formula of that particular series in order to get the simplest form of the given series so that the differentiation becomes easy. Then apply the rules of the differentiation in order to get the solution.

Complete step by step answer:
In this question we are given some kind of series or sequence. So our first step is to find which kind of series is given to us.
If the series is an AP series i.e. Arithmetic Progression then common difference between the consecutive numbers should be equal. So let us check the common differences in the series. The given series is \[y=1+\dfrac{1}{x}+\dfrac{1}{{{x}^{2}}}+\dfrac{1}{{{x}^{3}}}+.....\infty \]
To check the common difference formula used is, \[{{a}_{2}}-{{a}_{1}}={{a}_{4}}-{{a}_{1}}\]
Where,
\[{{a}_{1}}=\] first term of the series.
\[{{a}_{2}}=\] second term of the series.
\[{{a}_{3}}=\] third term of the series.
\[{{a}_{4}}=\] fourth term of the series.
From the given series,
\[\begin{align}
  & \Rightarrow {{a}_{2}}-{{a}_{1}}=\dfrac{1}{x}-1 \\
 & \Rightarrow {{a}_{2}}-{{a}_{1}}=\dfrac{1-x}{x} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow {{a}_{4}}-{{a}_{3}}=\dfrac{1}{{{x}^{3}}}-\dfrac{1}{{{x}^{2}}} \\
 & \Rightarrow {{a}_{4}}-{{a}_{3}}=\dfrac{1-x}{{{x}^{3}}} \\
\end{align}\]
We can say that the given series is not in AP because \[{{a}_{2}}-{{a}_{1}}\ne {{a}_{4}}-{{a}_{3}}\] .
Now we will check for the GP series. It is a series in which the common ration between the consecutive numbers should be equal. So now let us check the common ratio.
To check the common ratio formula used is, \[\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{4}}}{{{a}_{1}}}\]
Where,
\[{{a}_{1}}=\] first term of the series.
\[{{a}_{2}}=\] second term of the series.
\[{{a}_{3}}=\] third term of the series.
\[{{a}_{4}}=\] fourth term of the series.
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{\dfrac{1}{x}}{1} \\
 & \Rightarrow \dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{1}{x} \\
\end{align}\]
\[\begin{align}
  & \Rightarrow \dfrac{{{a}_{4}}}{{{a}_{3}}}=\dfrac{\dfrac{1}{{{x}^{4}}}}{\dfrac{1}{{{x}^{3}}}} \\
 & \Rightarrow \dfrac{{{a}_{4}}}{{{a}_{3}}}=\dfrac{{{x}^{3}}}{{{x}^{4}}} \\
 & \Rightarrow \dfrac{{{a}_{4}}}{{{a}_{3}}}=\dfrac{1}{x} \\
\end{align}\]
We can say that the given series is in GP because \[\dfrac{{{a}_{2}}}{{{a}_{1}}}=\dfrac{{{a}_{4}}}{{{a}_{1}}}\] .
So the given series is the infinite GP series using the sum formula of infinite GP, we get
Sum \[=\dfrac{a}{1-r}\]
Where,
\[a=\] first term
\[r=\] common ratio
By substituting the values we get,
\[\Rightarrow y=\dfrac{1}{1-\dfrac{1}{x}}\]
\[\Rightarrow y=\dfrac{1}{\dfrac{x-1}{x}}\]
\[\Rightarrow y=\dfrac{x}{x-1}\]
The infinite GP series is now converted into the simplest form so that the differentiation becomes easy.
Now differentiate the function \[y\]to get the answer. Use the quotient rule of the differentiate to solve
\[\Rightarrow y=\dfrac{x}{x-1}......(1)\]
\[\Rightarrow \dfrac{d}{dx}y=\dfrac{d}{dx}\dfrac{x}{x-1}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{(x-1)\times 1-x\times (1-0)}{{{(x-1)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{x-1-x}{{{(x-1)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{{{(x-1)}^{2}}}\]
In order to match from the options we have to convert the form to the answer.
Squaring on both the sides of equation \[(1)\], we get
\[\Rightarrow {{y}^{2}}=\dfrac{{{x}^{2}}}{{{(x-1)}^{2}}}\]
Substitute this value in \[\dfrac{dy}{dx}\] , we get
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{{{y}^{2}}}{{{x}^{2}}}\]

So, the correct answer is “Option D”.

Note: If we are given with some starting numbers of the series then to find the \[{{n}^{th}}\]term of the Geometric Progression or GP we will use the formula which is given as \[{{a}_{n}}=a{{r}^{n-1}}\], where \[a\] is the first term of the series, \[r\] is the common ratio between the consecutive numbers and \[n\] is the number of term.