Question

If we have an expression $y = {x^2} + \dfrac{1}{{{x^2} + \dfrac{1}{{{x^2} + \dfrac{1}{{{x^2} + .....\infty }}}}}}$, then $\dfrac{dy}{dx}$ is equal to
1). $\dfrac{{2xy}}{{\left( {2y - {x^2}} \right)}}$
2). $\dfrac{{xy}}{{\left( {y + {x^2}} \right)}}$
3). $\dfrac{{xy}}{{\left( {y - {x^2}} \right)}}$
4). $\dfrac{{2x}}{{\left( {2 + {x^2}/y} \right)}}$

Answer 1

First we will let the repeated value in the function as y. Then, in order to find $\dfrac{dy}{dx}$ we will differentiate the function with respect to x using the chain rule of differentiation. Then, we will separate the $\dfrac{dy}{dx}$ terms from the other terms so that we can find the value of $\dfrac{dy}{dx}$ and after simplifying the function we will get the required answer.

Complete step-by-step solution:
$y = {x^2} + \dfrac{1}{{{x^2} + \dfrac{1}{{{x^2} + \dfrac{1}{{{x^2} + .....\infty }}}}}}$
Same term is being repeated. So, we will let that whole term as y.
$y = {x^2} + \dfrac{1}{y}$
Now we will multiply the equation by y on both sides.
$\Rightarrow {y^2} = {x^2}y + \dfrac{y}{y}$
$\Rightarrow {y^2} = {x^2}y + 1$
Now, we will differentiate this function with respect to x.
$\Rightarrow 2y.\dfrac{{dy}}{{dx}} = {x^2}\dfrac{{dy}}{{dx}} + y.2x + 0$
$\Rightarrow 2y\dfrac{{dy}}{{dx}} = {x^2}\dfrac{{dy}}{{dx}} + 2xy$
Now, on L.H.S we will separate all the $\dfrac{dy}{dx}$ terms and on R.H.S. we will take the remaining terms.
$\Rightarrow 2y\dfrac{{dy}}{{dx}} - {x^2}\dfrac{{dy}}{{dx}} = 2xy$
$\Rightarrow \dfrac{{dy}}{{dx}}(2y - {x^2}) = 2xy$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2xy}}{{2y - {x^2}}}$
The value of $\dfrac{dy}{dx}$ is $\dfrac{{2xy}}{{2y - {x^2}}}$.
So, option (1) is the correct answer.

Note: In order to solve these types of questions the key is to differentiate and separate all the like terms of the equation to easily find the answer. Trigonometric properties are the basic chain rule of differentiation used to solve this problem.
The chain rule is a formula to compute the derivative of a composite function. It may be possible to apply the chain rule even when there are no formulas for the functions which are being differentiated. This can happen when the derivatives are measured directly.