Question

If $-2{{y}^{2}}-2+{{x}^{2}}=0$ then find $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at the point (-2 , 1) in simplest form.

Answer 1

The $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ means the second derivative of the function y with respect to x. Therefore, differentiate the given equation twice and with the help of suitable changes find the expression for $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$. Then find its value at (-2 , 1).

Formula used:
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$

Complete step by step answer:
The given equation in the question is $-2{{y}^{2}}-2+{{x}^{2}}=0$ ….. (i),
which has two variables, x and y. Let us assume that the variable y is a function that is dependent on the variable x.
Let us understand what is meant by $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$. $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ is the second derivative of the function y with respect to x.
Derivative of a function with respect to some variable is defined as the rate of change in that function with respect to the change in the independent variable.In other words, it tells us how the function y changes when the variable x changes. Let us differentiate each term of the equation (i) with respect to x.

With this we get that $\dfrac{d}{dx}\left( -2{{y}^{2}} \right)+\dfrac{d}{dx}\left( -2 \right)+\dfrac{d}{dx}\left( {{x}^{2}} \right)=\dfrac{d}{dx}(0)$ …. (i)
Derivative of a constant is zero. Therefore, $\dfrac{d}{dx}(-2)=\dfrac{d}{dx}(0)=0$.
And $\dfrac{d}{dx}\left( {{x}^{2}} \right)=2x$
By using the chain rule we get $\dfrac{d}{dx}\left( -2{{y}^{2}} \right)=-2\left( 2y \right)\dfrac{dy}{dx}=-4y\dfrac{dy}{dx}$
Substitute all the found values in equation (i).
$\Rightarrow -4y\dfrac{dy}{dx}+2x=0$
$\Rightarrow 2y\dfrac{dy}{dx}=x$ …. (ii)

Now, let us differentiate the above equation with respect to x again to find the expression for the second derivative of y. Then,
$\Rightarrow \dfrac{d}{dx}\left( 2y\dfrac{dy}{dx} \right)=\dfrac{d}{dx}x$ ….. (iii)
By using product rule we get that $\dfrac{d}{dx}\left( 2y\dfrac{dy}{dx} \right)=\dfrac{d}{dx}(2y)\dfrac{dy}{dx}+(2y)\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)$
$\Rightarrow \dfrac{d}{dx}\left( 2y\dfrac{dy}{dx} \right)=2\dfrac{dy}{dx}.\dfrac{dy}{dx}+(2y)\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)$
$\Rightarrow \dfrac{d}{dx}\left( 2y\dfrac{dy}{dx} \right)=2{{\left( \dfrac{dy}{dx} \right)}^{2}}+2y\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)$
And $\dfrac{d}{dx}x=1$
Substitute these values equation (iii)
$\Rightarrow 2{{\left( \dfrac{dy}{dx} \right)}^{2}}+2y\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=1$ …. (iv)

Now, from (ii) we get that $\dfrac{dy}{dx}=\dfrac{x}{2y}$
Substitute this in (iv)
$\Rightarrow 2{{\left( \dfrac{x}{2y} \right)}^{2}}+2y\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=1$
$\Rightarrow 2\left( \dfrac{{{x}^{2}}}{4{{y}^{2}}} \right)+2y\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=1$
On simplifying we get $2y\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=1-\dfrac{{{x}^{2}}}{2{{y}^{2}}}=\dfrac{2{{y}^{2}}-{{x}^{2}}}{2{{y}^{2}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{y}^{2}}-{{x}^{2}}}{2{{y}^{2}}}\times \dfrac{1}{2y}=\dfrac{2{{y}^{2}}-{{x}^{2}}}{4{{y}^{3}}}$
Now, substitute $x=-2,y=1$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2{{(1)}^{2}}-{{(-2)}^{2}}}{4{{(1)}^{3}}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{2-4}{4}\\\ \therefore\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\dfrac{1}{2}$

Therefore, the value of $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ at (-2,1) is $-\dfrac{1}{2}$.

Note: Sometimes students misunderstand the means of $\dfrac{dy}{dx}$. It is a notation for derivative or rate of change in y with respect to the change in x and not just a ratio between ‘dy’ and ‘dx’.Do not misunderstand that $\dfrac{dx}{dy}=\dfrac{1}{\left( \dfrac{dy}{dx} \right)}$. The derivative of x with respect to y will have a different meaning.