Can someone explain "implicit differentiation" and show how to use it?

Question

I seem not to be grasping the concept of "implicit differentiation". I already know the power rule, product rule, quotient rule, and chain rule.

For example, a problem in my calculus book asks me to find the derivative of x * y + 2x + 3x^2 = 4 by (a) setting the equation equal to "y", and (b) using implicit differentiation.

I already can solve part a of this problem, but i don't know how to do part b. Here's what I've got:

x * y + 2x + 3x^2 = 4
(d/dx)(x) * (d/dx)(y) + (d/dx)(2x) + (d/dx)(3x^2) = (d/dx)(4)
1 * (d/dx)(y) + 2 + 6x = 0

Then I tried to follow the book's example, and did:
(d/dx)(y) = (d/dy)(y) * (d/dx) = 1 * (d/dx)

So, how does this implicit differentiation work and what do I do now, considering I'm correct?

Answer 1

For a term like x*y, you need to use the product rule. You get

x*dy/dx + y*dx/dx = x*dy/dx + y.

You differentiated the rest of it correctly. So now you have

x*dy/dx + y + 2 + 6x = 0

which you can solve for dy/dx.

In general, you perform implicit differentiation by thinking of y as some function of x, you just don't know what it is (unless you solve for it like you did in part (a), but often you can't, in which case implicit differentiation is the only way to find dy/dx).

Just as you learned the other differentiation rules from how they're applied to an arbitrary function, such as d/dx[f(x)^n] = nf(x)^(n-1)*f'(x), you apply those same rules during implicit differentiation to an implicitly-defined function of x, that's not completely specified. As a result, you're frequently using the chain rule.

Example: implicitly differentiating

tan y = x

gives you

sec^2 y * dy/dx = 1,

so dy/dx = 1/sec^2 y.

Answer 2

You first need to understand implicit functions. They're equations that can be written in the form f(x,y)=0 and typically cannot be solved for y. For example -- x*y^3+x^2*y-5=0

In this case different segments of the graph would by themselves pass the vertical line test and by themselves express explicit functions. So in a sense an implicit function represents many explicit functions. We're just not typically clever enough to find them by solving for y.

Now ... even though we can't find the equation, we can find the slope of the tangent line at any point. This is what implicit differentiation does for us.

I think you're not really asking "what is implicit differentiation?" but rather "how do I use implicit differentiation for this problem?" So, here it is.

The assumption is that y is a function of x, but that we cannot find the explicit formula of y in terms of x. If you understand the chain rule, you should be able to do the following derivatives. I will abbreviate dy/dx as y'.

For example d/dx (y^2) = 2y * y'

Another example would be d/dx sin(y) = cos(y) * y'

Notice in each case the derivative is done with respect to y first, then multiplied by the derivative of y with respect to x (or y').

So if you got that, you should be able to do implicit differentiation using normal rules. The only difference is that with implicit differentiation, you'll have a bunch of y' floating around. You'll have to solve for y' algebraically.

Take for example x^2 + y^2 = 9. This implicit equation represents a circle or radius 3, centered at the origin. It describes two explicit functions which would be the top and bottom halves of the circle. To find the slope of the circle at ANY point, we'll take the derivative implicitly.

x^2 + y^2 =9

Taking the derivative with respect to x ...

d/dx (x^2 + y^2) = d/dx (9)
2x + 2y * y' = 0

Finally, solving for y' ... y' = -x/y

Your problem goes like this ...

x * y + 2x + 3x^2 = 4
d/dx (x * y + 2x + 3x^2) = d/dx (4)
y + x * y' + 2 + 6x = 0

Solving for y' you get ... y' = -(6x+y+2)/x

In each of these examples, you could solve for y and take the derivative as always. In each case, whether you find the explicit function first and then take the derivative or use the implicit formula and then take the derivative ... the two formulas will always evaluate to the same thing at a point on the curve. (you will get the same slope, but the formula will look different)

Answer 3

Wow. I'm impressed. I just stumbled onto this question and thought it a grammatical english question. I flunked math badly mostly because of lack of interest, mediocre instructors. I got a little tutoring once after the fact and found it to be what they say; a different way to say things, a pure science of delicate illusions. Dude, I don't believe in envy but I kinda do envy you. I guess I'll hazard a guess by saying that it is implied in the way a formula is written out that it's factors determining the result are arrived at only one way and this implied by same. Looks like algebra. Yeah I know. D -. I tried! ( LOL )