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.