Problems with implicit differentiation

[giant016]

In a problem where I need to use implicit diff. to find the slope of a line such as:

y^2 + x^2 = 9


2y (dy/dx) + 2x = 0


dy/dx = -y/x

Where does the (dy/dx) after the 2y (in the second part) come from? I've already differentiated y^2, x^2, and 9. Why isn't it just 2y + 2x = 0? I've looked all over, and in every problem an extra (dy/dx) or two just seem to pop up out of nowhere. Thanks!

 

[JonF]

Think of Y as a function in X. Then it's just the chain rule.

For example if y=e^x+x^3
y' = e^x + 3x^2

y^2 + x^2 = 9 is

(e^x+x^3)^2 + x^2 - 9 = 0
when we plug in y


differentiating this we get
2(e^x+x^3)(e^x + 3x^2) + 2x = 0

or 2y*(dy/dx) +2x = 0

when you do implicit differentiation you are doing it for a general y function of x instead of something specific (like as in our case of y=e^x+x^3)

 

[tacman]

The (dy/dx) after the 2y comes from the chain rule. When you differentiate y^2, you need to use the chain rule because y is a function of x. So the derivative of y^2 is 2y(dy/dx). Similarly, when you differentiate x^2, the derivative is 2x(dx/dx), but since dx/dx is just 1, it is usually not written. In this problem, you are essentially finding the derivative of y^2 + x^2 with respect to x, so you need to use the chain rule for both terms.

It is important to remember that when using implicit differentiation, you are treating y as a function of x, even though it is not explicitly written as y = f(x). So every time you differentiate y, you need to use the chain rule. This can be a source of confusion for some students, but with practice, it will become more natural.

In some problems, the (dy/dx) may not be explicitly written, but it is still present. For example, in the equation y^2 + x^2 = 9, if you were to solve for dy/dx, you would end up with dy/dx = -y/x. The (dy/dx) is implied in this case.

Overall, implicit differentiation can be tricky at first, but with practice and a good understanding of the chain rule, it becomes easier to handle. Just remember to always use the chain rule when differentiating y and keep an eye out for the (dy/dx) terms.