DifferentiatingDate: 01/14/98 at 11:53:53 From: Stephen Hack Subject: Differentiation Please help me differentiate (y^2) * cos (1/y) = 2x + 2y Date: 01/30/98 at 16:27:04 From: Doctor Joe Subject: Re: differentiation Dear Stephen, I shall not answer your question directly. I hope that you can do it yourself. What I shall do is to give you some advice while performing an example of this type of implicit differentiation. Recall that if f is a function of y and y is also a function of x, then to differentiate f with respect to x, we need to apply the so-called chain-rule, right? d d dy -- f(y) = (-- f(y))* -- dx dy dx which can also be written as: df df dy -- = -- * -- dx dy dx For instance, differentiating y^3*sin(1/y^2) with respect to x gives: {d/dy (y^3*sin(1/y^2))} * dy/dx = {3y^2 * sin(1/y^2) + y^3 * cos(1/y^2) * (-2/y^3)} * dy/dx In a more complicated example, the next thing you would do would be to group the terms that have a factor of dy/dx together. To show you another example: Differentiate with respect to x, x^2 + y^3 = y We start differentiating implicitly with respect to x: 2x + (3y^2 * dy/dx) = dy/dx => dy/dx * (3y^2 - 1) = -2x => dy/dx = 2x/(1-3y^2) After grouping the dy/dx terms together, we were able to isolate them on one side of the equation. Note that the final expression of dy/dx usually involves a quotient. In the last example that I gave, the denominator is (1-3y^2). Don't think that everything is fine. Sometimes you have to check to see whether your derivative makes sense at certain special points - in this case it is the value when y = sqrt(1/3) . Now you can solve the problem you posed using the same methods I illustrated in my examples. You will go through the following steps: 1) differentiate each term with respect to x (and use the chain-rule when a term has y's in it), 2) rearrange and group the terms to solve for dy/dx. Feel free to mail me if in doubt. Regards. -Doctor Joe, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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