How to Differentiate Implicitly
Sometimes you are asked to differentiate an equation that’s not solved for y, like y^{5} + 3x^{2 } = sin x – 4y^{3}. This equation defines y implicitly as a function of x, and you can’t write it as an explicit function because it can’t be solved for y. For such a problem, you need implicit differentiation. When differentiating implicitly, all the derivative rules work the same, with one exception: When you differentiate a term with a y in it, you use the chain rule with a little twist.
For example, you can use the chain rule to differentiate something like sin (x^{3}) as follows: start with the outside function, sin, and differentiate that, IGNORING what’s inside — in this case, x^{3}. To make sure you ignore the inside, temporarily replace the inside function with the word stuff. In this example, the derivative of sine is cosine, so the derivative of sin (stuff) is
You finish the problem by finding the derivative of the stuff, x^{3}, which is 3x^{2}, and then making the substitutions to give you
With implicit differentiation, a y works like the word stuff. Thus, because
The twist is that while the word stuff is temporarily taking the place of some known function of x (x^{3} in this example), y is some unknown function of x (you don’t know what the y equals in terms of x). And because you don’t know what y equals, the y and the
But the concept is exactly the same, and you treat y just like the stuff. You just can’t make the switch back to xs at the end of the problem like you can with a regular chain rule problem.
Here’s an example. Differentiate y^{5} + 3x^{2 } = sin x – 4y^{3}:

Differentiate each term on both sides of the equation.
y^{5} + 3x^{2 } = sin x – 4y^{3}
For the first and fourth terms, you use the power rule and, because these terms contain ys, you also use the chain rule. For the second term, you use the regular power rule. And for the third term, you use the regular sine rule.

Collect all terms containing a
on the left side of the equation and all other terms on the right side.

Factor out

Divide for the final answer.
Note that this derivative is expressed in terms of x and y instead of just x. So, if you want to evaluate the derivative to get the slope at a particular point, you need to have values for both x and y to plug into the derivative.
Also note that in many textbooks, the symbol
is used instead of
in every step of solutions like the one shown here. You may find
easier and less cumbersome to work with. But
does have the advantage of reminding you that you’re finding the derivative of y with respect to x. Either way is fine. Take your pick.