Implicit vs. Explicit DifferentiationDate: 01/04/2001 at 20:18:35 From: Meghan Fraser Subject: Implicit vs. Explicit Differentiation What is the exact definition of 'Implicit Differentiation'? I have a very hard time with math terms (please use layman's terms). What is 'Explicit Differentiation'? I don't even know what the words implicit and explicit mean! What is the difference between them? Which do you use when? Thank you for the help. Date: 01/05/2001 at 17:26:09 From: Doctor Fenton Subject: Re: Implicit vs. Explicit Differentiation Dear Meghan, Thanks for writing to Dr. Math. This is a good question, and it depends on remembering what we mean by a function. Recall that a function is a rule that assigns exactly one y-value to a given x. Another way of stating this is that a vertical line always intersects the graph of a function in only one point. A function defined in the form y = f(x) is said to be "explicitly" defined. That means that y = x^2 or y = sin(x) is explicitly defined. Sometimes we have a graph given in a form that isn't, or perhaps can't be, solved for y. For example: x^2 + y^2 = 1 or y*e^y = x In the first case, you probably recognize the equation as the equation of a circle centered at the origin of radius 1. The whole graph is not a function because vertical lines between x = -1 and x = 1 hit the circle twice. In this example, we actually can solve for y: y^2 = 1 - x^2 so either y = + sqrt(1 - x^2) or y = - sqrt(1 - x^2) Both of these formulas define explicit functions: y = + sqrt(1 - x^2) is the upper half of the circle; and the other formula defines the lower half of the circle. Each semicircle IS the graph of a function, because vertical lines now hit the graph only once. We say that a function graph that is part of a larger graph (which may not be a function graph in its entirety) is "implicitly" defined by the larger graph or its formula. The second example gives a graph that IS a function graph, but there is just no way to rewrite it in the form y = f(x), so we say in this case, too, that y is "implicitly" defined by the equation (y*e^y = x in this case). For implicit differentiation, we have a formula F(x,y) = 0 which we either can't or don't bother to solve for y. For example, the circle equation would be: x^2 + y^2 - 1 = 0 and the other equation would be: y*e^y - x = 0 We can still compute derivatives, by using the Chain Rule whenever y occurs in the formula. In the circle example, the left side is a sum of three functions of x: x^2, y^2, and the constant 1. The derivative of a sum is the sum of the derivatives, so we differentiate with respect to x (which I will denote by an apostrophe, ') : (x^2)' + (y^2)' + (1)' = (0)' The first term is easy: (x^2)' = 2x The derivative of a constant is 0, so 2x + (y^2)' + 0 = 0 Since we are thinking of y as being a function of x, even though we don't know its formula, we can use the Chain Rule: d(y^2) d(y^2) dy -------- = ------ * --- dx dy dx = 2y * y' (where I am now writing y' for dy/dx) The first term on the right is just 2y, because we are differentiating y^2 WITH RESPECT TO y, so our equation becomes: 2x + 2y*y' = 0 We now solve for y': 2y*y' = -2x -2x y' = ---- 2y -x y' = --- y Notice that y occurs in the answer. Remember how there were TWO functions defined by the circle equation: y1 = + sqrt(1 - x^2) and y2 = - sqrt(1 - x^2) We can differentiate these functions explicitly, and get -x -x y1' = ------------- and y2' = -------------- sqrt(1 - x^2) -sqrt(1 - x^2) or -x -x y1' = -- and y2' = -- y1 y2 (Instead of canceling the - signs in the second derivative, I have left them in, to show that the single formula from implicit differentiation computes BOTH derivatives with one computation!) For the second function, we would compute: (y*e^y)' = (x)' y'*(e^y) + y * (e^y)' = 1 (I used the Product Rule on the left) y'*e^y + y*(e^y * y') = 1 (since by the Chain Rule, d(e^y) d(e^y) dy ------ = ------ * -- dx dy dx = e^y * y' ) Solving for y' gives y' * (e^y + y*e^y) = 1 y' * e^y*(1 + y) = 1 1 y' = --------- e^y*(1+y) This answer again has the characteristic feature of implicit differentiation: y will appear in the answer. I hope this helps explain the idea. If you have further questions, please write again. - Doctor Fenton, The Math Forum http://mathforum.org/dr.math/ |
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