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.

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