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Question: For \[|x|<1,y=1+x+{{x}^{2}}+........\infty \], then \[\dfrac{dy}{dx}-y\] is equal to A. \[\dfrac{x...

For ∣x∣<1,y=1+x+x2+........∞|x|<1,y=1+x+{{x}^{2}}+........\infty , then dydx−y\dfrac{dy}{dx}-y is equal to
A. xy\dfrac{x}{y}
B. x2y2\dfrac{{{x}^{2}}}{{{y}^{2}}}
C. xy2\dfrac{x}{{{y}^{2}}}
D. xy2x{{y}^{2}}

Explanation

Solution

For solving this question, firstly we have to observe the given yy and check if we can simplify it, after that we have to find the derivative of yy with respect to xx and after finding the derivative we will substitute the values in dydx−y\dfrac{dy}{dx}-y and we will get our required answer.

Complete step by step answer:
Derivatives can be defined as the rate at which a quantity changes instantaneously to another quantity (i.e. varying rate of change of a given function with respect to any independent variable). It is used when there is any varying quantity. If we have to find the derivative of yy with respect to the xx means that we have to find the change in yy over the change in xx . In this yy is the dependent variable whereas xx is the independent variable. The process of finding the derivatives is known as the differentiation.
Derivative of any function can also be defined as the slope of the graph of any given function. Derivatives are classified on the basis of their order such as first order derivative and second order derivative.
First order derivative tells us whether the given function is increasing or decreasing. Whereas a second order derivative gives the idea about the shape of the graph (i.e. about the concavity of the graph), whether the graph is concave up or concave down.
We have given in the question: y=1+x+x2+........∞y=1+x+{{x}^{2}}+........\infty
As we can observe that the common ratio between the successive terms is xx so, we can say that this is an infinite geometric progression, in which a=1a=1 and r=xr=x
Sum of GP is given as:
sum=a1−xsum=\dfrac{a}{1-x}
So, y=11−xy=\dfrac{1}{1-x}
Now we have to find the derivative of yy with respect to the xx
dydx=ddx[11−x]\dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \dfrac{1}{1-x} \right]
⇒dydx=[(1−x)×0−(1×−1)](1−x)2\Rightarrow \dfrac{dy}{dx}=\dfrac{\left[ (1-x)\times 0-(1\times -1) \right]}{{{(1-x)}^{2}}}
⇒dydx=0+1(1−x)2\Rightarrow \dfrac{dy}{dx}=\dfrac{0+1}{{{(1-x)}^{2}}}
⇒dydx=1(1−x)2\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{{{(1-x)}^{2}}}
Now to find the value of dydx−y\dfrac{dy}{dx}-y , we will substitute the above values in this and it becomes as:
So, dydx−y=[1(1−x)2]−[1(1−x)]\dfrac{dy}{dx}-y=\left[ \dfrac{1}{{{(1-x)}^{2}}} \right]-\left[ \dfrac{1}{(1-x)} \right]
⇒[1−(1−x)(1−x)2]\Rightarrow \left[ \dfrac{1-(1-x)}{{{(1-x)}^{2}}} \right]
⇒[1−1+x(1−x)2]\Rightarrow \left[ \dfrac{1-1+x}{{{(1-x)}^{2}}} \right]
⇒[x(1−x)2]\Rightarrow \left[ \dfrac{x}{{{(1-x)}^{2}}} \right]
As from above calculations we know that: y=1(1−x)2y=\dfrac{1}{{{(1-x)}^{2}}}
Therefore, dydx−y=xy2\dfrac{dy}{dx}-y=x{{y}^{2}}

So, the correct answer is “Option D”.

Note: Derivatives are used in a variety of different fields. Economists use derivatives to describe the profit and loss in a business. Electrical engineers use derivatives to describe the change in current within an electric circuit. It can also be used to describe the motion of many different objects.