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Trying to self-learn calculus but stuck on this idea

  1. Dec 2, 2014 #1
    So I started with limits and everything went well, and then into derivatives, and a problem started.

    First I know how to find the derivative of basic things using like ( limit definition of derivative) or (rules for derivatives) or ( product rule/ quotient rule/ chain rule).....

    My problem probably falls under the " Implicit Differentiation" category

    But the problem is when the derivative becomes more than just D/DX, it becomes like DY/DX, and the videos/textbooks tell me stuff like : derivative of something WITH RESPECT to something... And the equation has more than x like x+y = something... I am trying to get this concept into my head but it is not clicking!
    Especially those questions that leave d/dy attached to them...

    What is the derivative of x with respect to y ?? What is the derivative of y with respect to x ? Why is there "respect"?
  2. jcsd
  3. Dec 2, 2014 #2


    Staff: Mentor

    The "with repect to" part indicates which variable you're considering to be the independent variable. As an example, let's start with the equation x = y2 - 3y. Here we have x in terms of y, so x is dependent on y, making y the independent variable.

    If we differentiate both sides of this equation, with respect to y, we get d/dy(x) = d/dy(y2 - 3y) = 2y - 3. The upshot is that dx/dy, the derivative of x with respect to y, is 2y - 3.

    Here's a more complicated problem, with the equation xy - y2 = 3x. I can differentiate each side with respect to either x or y. If I differentiate with respect to x, I get d/dx(xy - y2) = d/dx(3x), or x dy/dx + y dx/dx - 2y dy/dx = 3. This simplifies to x dy/dx + y - 2y dy/dx = 3. The first two terms come from applying the product rule when differentiating xy and noting that dx/dx = 1 (the derivative of a variable with respect to itself is just 1). The third term comes from applying the chain rule.

    If I want an explicit form for dy/dx, I can move the terms that don't involve dy/dx to the other side of the equation, and solve algebraically for dy/dx.

    On the other hand, I can differentiate the equation with respect to y, which involves similar operations. I leave this as an exercise for you.
  4. Dec 2, 2014 #3
    The big thing to understand here is that implicit differentiation kind of depends on the chain rule.

    First, derivatives with respect to something means that if y = 3(x^2), y = 3u, and u = x^2,
    then derivative of y with respect to u is 3
    and derivative of u with respect to x is 2x
    so chain rule (you know this, of course) 2x * 3 = derivative of y with respect to x = 6x (duh)

    so say that you need to find the derivative of x^2+y^2 = 2 with RESPECT TO X (classic example) take the derivative of each adding part (forgot basic math words)

    derivative of x^2 with RESPECT TO X is 2x (duh)

    derivative of y^2 with RESPECT TO X (not y) is 2y * (dy/dx); (because chain rule,,, its the derivative of the outside with respect to y (2y) TIMES the
    derivative of the inside (y) with respect to x (dy/dx)

    derivative of 2 = 0 (duh also)

    so now we have 2x + 2y * dy/dx = 0
    subtract 2x >>>> 2y * dy/dx = - 2x
    divide by 2y >>>> dy/dx = -x/y
    you now have the derivative.
  5. Dec 3, 2014 #4
    yeah, its basically the same thing as the chain rule.
    just remember when we say "a with respect to b" it always means a is the output(dependent variable) and b is the input(independent variable)
  6. Dec 3, 2014 #5
    If you have your x and y axes, the differentiating with respect to x is marching x forward a small amount in the horizontal direction and following along the curve to see how much y goes up by. Differentiating by y would be marching y up in the vertical direction a small amount and seeing how far along the curve x is.
  7. Dec 5, 2014 #6


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    Science Advisor

    When you are working with functions of a single variable, say f(x), then it is obvious that the value of f can only change when x changes and the derivative is the "rate of change" of f as x changes. If you have a function of more than one variable, say f(x, y), then f can change when either x or y changes. The "partial derivative with respect to x" is the rate of change of the value of f when only x changes. Similarly, the "partial derivative with respect to y" is the rate of change of the value of f when only y changes. Of course, it is possible for both x and y to change and then the rate of change of the value of f depends on both. That is the "directional derivative" and its value depends up the particular direction, in the xy-plane in which the (x, y) value is moving.
  8. Dec 5, 2014 #7
    I had some trouble with implicits myself. This guy on youtube was a lot of help. Get out your pencil and paper and get ready to take notes and work examples.
    He is worth the time it takes to go through the problems though. I hope it helps. I'm not very good at explaining in plain english what differentiation is. I just do it after I watch it done. http://www.profrobbob.com/calculus/implicit-differentiation
  9. Dec 7, 2014 #8
    The idea behind it is that, when we differentiate a function, we're sort of 'secretly' using the chain rule. Sometimes, we have equations that are explicitly defined as a function of ##x##, like:
    ##y = 5x - 2## (1)
    Of course, we could rearrange it so that it is 'implicitly' defined as a function of ##x##:
    ##-5x + y = -2## (2)
    Now, suppose you had (2) and didn't know you could rearrange it into (1). Using the chain rule (i.e., implicitly diferentiating), we get:
    ##\frac{d}{dx} (-5x + y) = \frac{d}{dx}-2##
    ##\frac{d(-5x)}{dx} + \frac{d(y)}{dx}) = \frac{d(-2)}{dx}##
    ##-5 + \frac{dy}{dx} = 0##
    ##\frac{dy}{dx} = 5##
    Of course, if we just differentiated (1), we'd get the same thing:
    ##\frac{d}{dx}(y = 5x - 2)##
    ##\frac{d(y)}{dx} = \frac{d(5x)}{dx} + \frac{d(-2)}{dx}##
    ##\frac{dy}{dx} = 5##

    Or, suppose we had the function:
    ##y^2 = 3x + 2##
    We could implicitly differentiate to get:
    ##\frac{d}{dx}y^2 = \frac{d}{dx}3x + \frac{d}{dx}2##
    ##2y \frac{dy}{dx} = 3##
    ##\frac{dy}{dx} = \frac{3}{2y}##
    Another way we could have done it is to rearrange the original equation into an explicit function of ##x##:
    ##y^2 = 3x + 2## ##\rightarrow## ##y = \sqrt{3x + 2}##
    Then, differentiating gives us:
    ##\frac{dy}{dx} = \frac{3}{2 \sqrt{3x + 2}}##
    But, ##\sqrt{3x+2} = y##, so:
    ##\frac{dy}{dx} = \frac{3}{2y}##, the same answer.
    An easy way to conceptualize the use of the chain rule when differentiating the ##y##-terms is to think of the ##y## as the argument of a function of ##x##. So, by the chain rule, if (for example),
    ##f = x^2##
    ##g = y##
    Then, ##f(g) = y^2##, so ##\frac{d}{dx}f(g) = 2(y)\cdot \frac{dy}{dx}##
    Or, basically,
    ##f = \mbox{(thing)}^n##
    ##g = \mbox{(another thing)}^m##
    ##\frac{d}{d\mbox{thing}}(f(g))= n(\mbox{another thing})^{n-1} \cdot m(\mbox{another thing})^{m-1}##
  10. Dec 7, 2014 #9


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    I'm not sure if this part has been answered above. When you have x + y = 5, you have restricted the combinations of x and y that satisfy the equation. These are implicit restrictions on x and y. You could say that y is defined by x because x = 1 forces y = 4. So y is a function of x, y(x). But you could also say that y defines x because y=0.5 forces x = 4.5. So x is a function of y, x(y).

    If a problem asks for the derivative of "y with respect to x", it wants you to consider y as a function of x. So we have x + y(x) = 5.
    If a problem asks for the derivative of "x with respect to y", it wants you to consider x as a function of y. So we have x(y) + y = 5.

    In this simple example for y with respect to x, it is easy to explicitly solve for y(x) and take the derivative.
    y(x) = 5-x
    dy/dx = -1.
    There can be much more complicated examples where solving for y(x) is not easy and it's easier to take the derivative of both sides of the original equation with respect to x. This leaves y(x) implicitly defined by the original equation and it uses implicit differentiation.
    d/dx( x + y(x) ) = d/dx 5
    1 + dy/dx = 0
    dy/dx = -1
  11. Dec 10, 2014 #10
    Thanks everyone for the amazing replies! I have to stop self-studying calculus for a while now that exams are here, but I will definitely check them out after the exams are done and try to get the most out of them.
  12. Dec 11, 2014 #11
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