# How do you differentiate 2x^2 = 15,000 - p^2?

Oct 24, 2015

Differentiate with respect to which variable?

#### Explanation:

I wouldn't ordinarily ask, but this looks like the kind of example used for a demand equation, where we might indeed want to know the rate of change of demand (x) with respect to price (p) oreven the relative rates of change with respect ti time (t).

With respect to x

$2 {x}^{2} = 15 , 000 - {p}^{2}$

$\frac{d}{\mathrm{dx}} \left(2 {x}^{2}\right) = \frac{d}{\mathrm{dx}} \left(15 , 000 - {p}^{2}\right)$

$4 x = - 2 p \frac{\mathrm{dp}}{\mathrm{dx}}$ $\text{ }$ (implicit differentiation is the chain rule)

$\frac{\mathrm{dp}}{\mathrm{dx}} = - \frac{2 x}{p}$

With respect to p

$2 {x}^{2} = 15 , 000 - {p}^{2}$

$\frac{d}{\mathrm{dp}} \left(2 {x}^{2}\right) = \frac{d}{\mathrm{dp}} \left(15 , 000 - {p}^{2}\right)$

$4 x \frac{\mathrm{dx}}{\mathrm{dp}} = - 2 p$ $\text{ }$ (implicit differentiation is the chain rule)

$\frac{\mathrm{dx}}{\mathrm{dp}} = - \frac{p}{2 x}$

With respect to t

$2 {x}^{2} = 15 , 000 - {p}^{2}$

$\frac{d}{\mathrm{dt}} \left(2 {x}^{2}\right) = \frac{d}{\mathrm{dt}} \left(15 , 000 - {p}^{2}\right)$

$4 x \frac{\mathrm{dx}}{\mathrm{dt}} = - 2 p \frac{\mathrm{dp}}{\mathrm{dt}}$ $\text{ }$ (implicit differentiation is the chain rule)

$2 x \frac{\mathrm{dx}}{\mathrm{dt}} = - p \frac{\mathrm{dp}}{\mathrm{dt}}$