# Implicit Differentiation

## Key Questions

• Let us find $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$ for ${x}^{3} + {y}^{3} = 1$.

First, let us find $\frac{\mathrm{dy}}{\mathrm{dx}}$.
${x}^{3} + {y}^{3} = 1$
by differentiating with respect to $x$,
$R i g h t a r r o w 3 {x}^{2} + 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$
by subtracting $3 {x}^{2}$,
$R i g h t a r r o w 3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {x}^{2}$
by dividing by $3 {y}^{2}$,
$R i g h t a r r o w \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{x}^{2}}{{y}^{2}}$

Now, let us find $\frac{{d}^{2} y}{{\mathrm{dx}}^{2}}$.
by differentiating with respect to $x$,
Rightarrow{d^2y}/{dx^2}=-{2x cdot y^2-x^2 cdot 2y{dy}/{dx}}/{(y^2)^2} =-{2x(y^2-xy{dy}/{dx})}/{y^4}
by plugging in $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{x}^{2}}{{y}^{2}}$,
$R i g h t a r r o w \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2 x \left[{y}^{2} - x y \left(- {x}^{2} / {y}^{2}\right)\right]}{y} ^ 4 = - \frac{2 x \left({y}^{2} + {x}^{3} / y\right)}{y} ^ 4$
by multiplying the numerator and the denominator by $y$,
$R i g h t a r r o w \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2 x \left({y}^{3} + {x}^{3}\right)}{y} ^ 5$
by plugging in ${y}^{3} + {x}^{3} = 1$,
$R i g h t a r r o w \frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{2 x}{y} ^ 5$

• As the first step, we will differentiate once, and apply the product rule:

$\frac{d}{\mathrm{dx}} \left[{x}^{3}\right] \cdot {y}^{3} + \frac{d}{\mathrm{dx}} \left[{y}^{3}\right] \cdot {x}^{3} = \frac{d}{\mathrm{dx}} \left[8\right]$

For ${y}^{3}$, remember to use the chain rule. Simplifying yields:

$3 {x}^{2} {y}^{3} + 3 {y}^{2} {x}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

Now, we will solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 {x}^{2} {y}^{3}}{3 {y}^{2} {x}^{3}}$

We can cancel off the 3, an ${x}^{2}$, and a ${y}^{2}$, which will yield:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x}$

Now, differentiate once again. We will apply the quotient rule:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} = - \frac{x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - y \cdot 1}{x} ^ 2$

Looking back at the previous equation for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we can substitute into our equation for the second derivative to get it in terms of only $x$ and $y$:

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

Simplifying yields:

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

• Implicit differentiation is a way of differentiating when you have a function in terms of both x and y. For example:

${x}^{2} + {y}^{2} = 16$

This is the formula for a circle with a centre at (0,0) and a radius of 4

So using normal differentiation rules ${x}^{2}$ and 16 are differentiable if we are differentiating with respect to x

$\frac{d}{\mathrm{dx}} \left({x}^{2}\right) + \frac{d}{\mathrm{dx}} \left({y}^{2}\right) = \frac{d}{\mathrm{dx}} \left(16\right)$

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

To find $\frac{d}{\mathrm{dx}} \left({y}^{2}\right)$ we use the chain rule:

$\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

Rearrange for $\frac{\mathrm{dy}}{\mathrm{dx}}$

dy/dx=(-2x)/(2y

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{y}$

So essentially to use implicit differentiation you treat y the same as an x and when you differentiate it you multiply be $\frac{\mathrm{dy}}{\mathrm{dx}}$