When dealing with problems where you need to differentiate implicit functions, think of y as y(x). This helps when differentiating and tells us that we will need to use the chain rule. Let’s investigate how to do the implicit differentiation process on a function like the one given above, where y is on both sides. Functions like this make it difficult to explicitly express it only in terms of y.

Step 1:

First, we will differentiate both sides with respect to x. 

[latex]\large{\frac{d}{dx}[x^3] + \frac{d}{dx}[y^3] = \frac{d}{dx}[6xy]}[/latex]

The first term is easy enough for us to find the derivative of. Simply using the power rule, we know the derivative is [latex]3x^2[/latex]. What about the second term on the left side and also the term on the right? Let’s take a look at them one by one.

Remember, since we are treating y as y(x), we can say:

[latex]\large{\frac{d}{dx}[y^3] = \frac{d}{dx}[[y(x)]^3]}[/latex]

Hence, using our knowledge of the power rule and chain rule, we have:

[latex]\large{\frac{d}{dx}[y(x)^3] = 3[y(x)]^2 \cdot \frac{dy}{dx} = 3y^2y’}[/latex].

Notice how we replaced the y(x) back with y since we only put it that way to visually help us when differentiating. Also note that [latex]\frac{dy}{dx}[/latex] is the same thing as [latex]y'[/latex]. 

From now on, we will mostly use [latex]y'[/latex], as it makes the overall solution look cleaner.

Now let’s take a look at the right side. If you look carefully, you will notice that we are taking the derivative of the product of two functions, which are [latex]6x[/latex] and [latex]y[/latex]. Remember we are treating y as a function of x.

In this case, we can use the product rule to find the derivative, as shown:

[latex]\large{\frac{d}{dx}[6xy] = 6xy’ + 6y }[/latex].

Now, combining everything, we have:

[latex]\large{3x^2 + 3y^2y’ = 6xy’ + 6y}[/latex]

Step 2:

The goal is to solve for the derivative [latex]\frac{dy}{dx}[/latex] or [latex]y'[/latex]. So now, we will perform some simple algebra to solve for it, as shown:

[latex]\large{3y^2y’ – 6xy’ = 6y – 3x^2}[/latex]

[latex]\large{y'(3y^2-6x) = 6y-3x^2}[/latex]

[latex]\large{y’ = \frac{6y-3x^2}{3y^2-6x}}[/latex]

Step 3:

We can factor the 3 out from both the numerator and denominator, which will give us the final answer for the derivative:

[latex]\large{y’ = \frac{3(2y-x^2)}{3(y^2-2x)}}[/latex]

[latex]\large{y’ = \frac{2y-x^2}{y^2-2x}}[/latex]

That is how you use implicit differentiation to solve for the derivative of implicit functions. Try out more practice problems for yourself!