Tutorials on Advanced Math and Computer Science Concepts

# Implicit Differentiation

So far, we have seen functions that are easy to isolate for one variable. We typically have functions in the form of f(x) = [stuff] or y = [stuff]

There are many relations and functions that can't be rearranged in this way. Take for example the equation of a circle, $x^2 + y^2 = r^2$. In this case, we want to be able to find the rate of change, but it isn't easy to isolate for y. In cases like these, implicit differentiation is useful.

Example: Find the derivative of $x^2 + y^2 = 25$ using implicit differentiation

If we start calculating the derivative of this function, here is what we get:

$\frac{dy}{dx}(x^2) + \frac{dy}{dx}(y^2) = \frac{dy}{dx}(25)$
$\Longrightarrow 2x + \frac{dy}{dx}(y^2) = 0$

At this point, we need to discuss how we can compute $\frac{dy}{dx}(y^2)$. Since y is a function of x, we can't just compute the derivative as if it were $x^2$. Instead, we treat it like a chain rule problem. We take the derivative of the y term, then multiply the derivative of y with respect to x. Doing so gives us:

$\frac{dy}{dx}(y^2) = 2y * \frac{dy}{dx}$

This would make our equation:

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

Remember, our goal is to determine what $\frac{dy}{dx}$ is equal to. Since this is the case, we simply need to rearrange the equation above to isolate for $\frac{dy}{dx}$.

$2x + 2y * \frac{dy}{dx} = 0$
$\Longrightarrow 2y * \frac{dy}{dx} = -2x$
$\Longrightarrow \frac{dy}{dx} = \frac{-2x}{2y}$
$\Longrightarrow \frac{dy}{dx} = \frac{-x}{y}$

Implicit differentiation is quite easy once you get the hang of it. Every problem will run the same way. We take the derivatives, multiply $\frac{dy}{dx}$ by every y term, then solve for $\frac{dy}{dx}$. Let's try a few more examples.

Example: Determine y' for $x^3 + y^2 = 3xy$

$\frac{dy}{dx}(x^3) + \frac{dy}{dx}(y^2) = \frac{dy}{dx}(3xy)$
$\Longrightarrow 3x^2 + 2y * \frac{dy}{dx} = 3y+3x * \frac{dy}{dx}$

Notice that the 3xy term uses product rule. We still treat the y terms as regular functions, so derivative rules still apply. The only difference is that the y terms get the $\frac{dy}{dx}$ tacked on.

$3x^2 - 3y = 3x * \frac{dy}{dx} - 2y * \frac{dy}{dx}$
$\Longrightarrow 3x^2 - 3y = \frac{dy}{dx} * (3x-2y)$
$\Longrightarrow \frac{3x^2 - 3y}{3x - 2y} = \frac{dy}{dx}$

As you can see, implicit differentiation allows us to take derivatives of some new functions. Among them is the inverse trigonometric functions.

Example: Given $y = sin^{-1}(x)$, determine y'

To be able to find this derivative, we need to rearrange the equation. Recall that if $y = sin^{-1}(x)$, then $sin(y) = x$. Knowing this, we can apply implicit differentiation.

$\frac{dy}{dx}(sin(y)) = \frac{dy}{dx}(x)$
$\Longrightarrow cos(y) \frac{dy}{dx} = 1$
$\Longrightarrow \frac{dy}{dx} = \frac{1}{cos(y)}$

So now, all we need to do is write cos(y) in terms of x and we are done. We stated that $sin(y) = x$, so if we can rewrite cos(y) as sin(y), we can replace it with x. To do this, we use Pythagorean's identity.

$cos^2(y) + sin^2(y) = 1$
$\Longrightarrow cos^2(y) = 1 - sin^2(y)$
$\Longrightarrow cos(y) = \sqrt{1 - sin^2(y)}$

Therefore, we can substitute cos(y) to get $\frac{1}{1-sin^2(y)}$. Since $sin^2(y) = x^2$, we get:

$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$, which is our solution.