# How do you find the second derivative for the implicit equation x^2+y^2 = a^2?

Jun 29, 2017

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - {a}^{2} / {y}^{3}$

#### Explanation:

We have:

${x}^{2} + {y}^{2} = {a}^{2}$

This represents a circle of radius $a$ centred on the origin.

If we differentiate implicitly wrt $x$ we get:

$2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$\therefore x + y \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

Now, differentiating implicitly again, and applying the product rule, we get:

$\therefore 1 + \left(y\right) \left(\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}\right) + \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$\therefore 1 + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 0$

$\therefore 1 + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {\left(- \frac{x}{y}\right)}^{2} = 0$

$\therefore 1 + y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {x}^{2} / {y}^{2} = 0$

$\therefore y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + \frac{{x}^{2} + {y}^{2}}{y} ^ 2 = 0$

$\therefore y \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} + {a}^{2} / {y}^{2} = 0$

$\therefore \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - {a}^{2} / {y}^{3}$

Jun 29, 2017

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - {a}^{2} / {y}^{3.}$

#### Explanation:

${x}^{2} + {y}^{2} = {a}^{2.}$

$\therefore \frac{d}{\mathrm{dx}} \left({x}^{2} + {y}^{2}\right) = \frac{d}{\mathrm{dx}} \left({a}^{2}\right) = 0.$

$\therefore \frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 0.$

$\therefore 2 x + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = 0. \ldots \ldots \ldots \ldots \left(\ast\right) .$

Here, by the Chain Rule, we see that,

$\frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dy}} \left({y}^{2}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 2 y \frac{\mathrm{dy}}{\mathrm{dx}} .$

$\therefore \left(\ast\right) \Rightarrow 2 x + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0 , \mathmr{and} , x + y \frac{\mathrm{dy}}{\mathrm{dx}} = 0. \ldots \left({\ast}_{1}\right) .$

To get the ${2}^{n d}$ order deri., we rediff. this eqn. w.r.t. $x$, & get,

$1 + \frac{d}{\mathrm{dx}} \left(y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0. \ldots \ldots \ldots . . \left(\star\right) .$

Here, for $\frac{d}{\mathrm{dx}} \left(y \frac{\mathrm{dy}}{\mathrm{dx}}\right) ,$ we use the Product Rule :

$\frac{d}{\mathrm{dx}} \left(y \frac{\mathrm{dy}}{\mathrm{dx}}\right) = y \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(\frac{d}{\mathrm{dx}} \left(y\right)\right) ,$

$= y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) , i . e . , y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2.}$

$\therefore \left(\star\right) \Rightarrow 1 + y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2} = 0. \ldots . . \left({\star}_{1}\right) .$

But, $\left({\ast}_{1}\right) \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y} .$

$\therefore \left({\star}_{1}\right) \Rightarrow 1 + y \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + {\left(- \frac{x}{y}\right)}^{2} = 0.$

$\Rightarrow {y}^{2} + {y}^{3} \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + {x}^{2} = 0.$

$\Rightarrow {y}^{3} \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - \left({x}^{2} + {y}^{2}\right) , i . e . ,$

$\because , {x}^{2} + {y}^{2} = {a}^{2} , \therefore , \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = - {a}^{2} / {y}^{3.}$

Alternatively, the same result can be obtained by diff.ing, w.r.t.

$x$, the eqn. $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y} .$

Enjoy Maths.!