# How do you implicitly differentiate 5y^2=e^(xy)-3x ?

Mar 2, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y {e}^{y x} + 3}{10 y - x {e}^{x y}}$

#### Explanation:

For implicit differentiation we should remember that:

$\frac{d}{\mathrm{dx}} \left\{f \left(x\right)\right\} = f ' \left(x\right)$ and $\frac{d}{\mathrm{dy}} \left\{f \left(y\right)\right\} = \frac{\mathrm{dy}}{\mathrm{dx}} f ' \left\{y\right\}$
We can differentiate functions of $x$ normally but when differentiating functions of $y$ we must remember to apply the chain rule and pull out $\frac{\mathrm{dy}}{\mathrm{dx}}$.

In this case we just differentiate term by term along the equation:

$5 {y}^{2} = {e}^{x y} - 3 x$

The derivative of the first term $5 {y}^{2}$ is simply $10 y \frac{\mathrm{dy}}{\mathrm{dx}}$

For the second term: ${e}^{x y}$ we need to find the derivative of the exponent and then that multiply by the original function get the derivative.

So, to differentiate $x y$ we use the product rule to get:

$y + x \frac{\mathrm{dy}}{\mathrm{dx}}$

Now multiply that by the original expression to get:

$\left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{x y}$

And finally the last term: $3 x$ just differentiates to $3$.

So differentiating the equation gives us:

$10 y \frac{\mathrm{dy}}{\mathrm{dx}} = \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) {e}^{x y} + 3$

Now we do a bit of re arranging to det $\frac{\mathrm{dy}}{\mathrm{dx}}$ on one side of the equation:

$10 y \frac{\mathrm{dy}}{\mathrm{dx}} = y {e}^{y x} + x \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x y} + 3$

$10 y \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{y x} = y {e}^{y x} + 3$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(10 y - x {e}^{y x}\right) = y {e}^{y x} + 3$

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y {e}^{y x} + 3}{10 y - x {e}^{x y}}$