Cracking the AP Calculus BC Exam, 2016 Edition

HOW TO DO IT

By now, it should be easy for you to take the derivative of an equation such as y = 3x⁵ – 7x. If you’re given an equation such as y² = 3x⁵ – 7x, you can still figure out the derivative by taking the square root of both sides, which gives you y in terms of x. This is known as finding the derivative explicitly. It’s messy, but possible.

If you have to find the derivative of y² + y = 3x⁵ – 7x, you don’t have an easy way to get y in terms of x, so you can’t differentiate this equation using any of the techniques you’ve learned so far. That’s because each of those previous techniques needs to be used on an equation in which y is in terms of x. When you can’t isolate y in terms of x (or if isolating y makes taking the derivative a nightmare), it’s time to take the derivative implicitly.

Implicit differentiation is one of the simpler techniques you need to learn to do in calculus, but for some reason it gives many students trouble. Suppose you have the equation y² = 3x⁵ – 7x. This means that the value of y is a function of the value of x. When we take the derivative, , we’re looking at the rate at which y changes as x changes. Thus, given y = x² + x, when we write

= 2x + 1

we’re saying that “the rate” at which y changes, with respect to how x changes, is 2x + 1.

Now, suppose you want to find . As you might imagine

So here, . But notice that this derivative is in terms of x, not y, and you need to find the derivative with respect to y. This derivative is an implicit one. When you can’t isolate the variables of an equation, you often end up with a derivative that is in terms of both variables.

A fraction that has the same term in its numerator and denominator is equal to 1, so we write

= 2x(1) + 1(1) = 2x + 1

Every time we take a derivative of a term with x in it, we multiply by the term , but because this is 1, we ignore it. Suppose however, that we wanted to find out how y changes with respect to t (for time). Then we would have

If we wanted to find out how y changes with respect to r, we would have

and if we wanted to find out how y changes with respect to y, we would have

This is how we really do differentiation. Remember the following:

When you have an equation of x in terms of y, and you want to find the derivative with respect to y, simply differentiate. But if the equation is of y in terms of x, find and take its reciprocal to find . Go back to our original example.

y² + y = 3x⁵ – 7x

Notice how each variable is multiplied by its appropriate . Now, remembering that = 1, rewrite the expression this way: = 15x⁴ − 7.

Next, factor out of the left-hand side: (2y + 1) = 15x⁴ − 7.

Isolating gives you .

This is the derivative you’re looking for. Notice how the derivative is defined in terms of y and x. Up until now, has been strictly in terms of x. This is why the differentiation is “implicit.”

Confused? Let’s do a few examples and you will get the hang of it.

Example 1: Find if y³ – 4y² = x⁵ + 3x⁴.

Using implicit differentiation, you get

Remember that = 1: (3y² − 8y) = 5x⁴ + 12x³.

After you factor out , divide both sides by 3y² – 8y.

Note: Now that you understand that the derivative of an x term with respect to x will always be multiplied by , and that = 1, we won’t write anymore. You should understand that the term is implied.