# What is the implicit derivative of 4= (x+y)^2 ?

Apr 19, 2016

You can use calculus and spend a few minutes on this problem or you can use algebra and spend a few seconds, but either way you'll get $\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$.

#### Explanation:

Begin by taking the derivative with respect to both sides:
$\frac{d}{\mathrm{dx}} \left(4\right) = \frac{d}{\mathrm{dx}} {\left(x + y\right)}^{2}$

On the left, we have the derivative of a constant - which is just $0$. That breaks the problem down to:
$0 = \frac{d}{\mathrm{dx}} {\left(x + y\right)}^{2}$

To evaluate $\frac{d}{\mathrm{dx}} {\left(x + y\right)}^{2}$, we need to use the power rule and the chain rule:
$\frac{d}{\mathrm{dx}} {\left(x + y\right)}^{2} = \left(x + y\right) ' \cdot 2 {\left(x + y\right)}^{2 - 1}$
Note: we multiply by $\left(x + y\right) '$ because the chain rule tells us we have to multiply the derivative of the whole function (in this case ${\left(x + y\right)}^{2}$ by the inside function (in this case $\left(x + y\right)$).
$\frac{d}{\mathrm{dx}} {\left(x + y\right)}^{2} = \left(x + y\right) ' \cdot 2 \left(x + y\right)$

As for $\left(x + y\right) '$, notice that we can use the sum rule to break it into $x ' + y '$. $x '$ is simply $1$, and because we don't actually know what $y$ is, we have to leave $y '$ as $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$\frac{d}{\mathrm{dx}} {\left(x + y\right)}^{2} = \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(2 \left(x + y\right)\right)$

Now that we've found our derivative, the problem is:
$0 = \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(2 \left(x + y\right)\right)$

Doing some algebra to isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$, we see:
$0 = \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left(2 x + 2 y\right)$
$0 = 2 x + \frac{\mathrm{dy}}{\mathrm{dx}} 2 x + \frac{\mathrm{dy}}{\mathrm{dx}} 2 y + 2 y$
$0 = x + \frac{\mathrm{dy}}{\mathrm{dx}} x + \frac{\mathrm{dy}}{\mathrm{dx}} y + y$
$- x - y = \frac{\mathrm{dy}}{\mathrm{dx}} x + \frac{\mathrm{dy}}{\mathrm{dx}} y$
$- x - y = \frac{\mathrm{dy}}{\mathrm{dx}} \left(x + y\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- x - y}{x + y}$

Interestingly, this equals $- 1$ for all $x$ and $y$ (except when $x = - y$). Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$. We could have actually figured this out without using any calculus at all! Look at the equation $4 = {\left(x + y\right)}^{2}$. Take the square root of both sides to get $\pm 2 = x + y$. Now subtract $x$ from both sides, and we have $y = \pm 2 - x$. Remember these from algebra? The slope of this line is $- 1$, and since the derivative is the slope, we could have just said $\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$ and avoided all that work.