Slope of root x. . .Date: Wed, 14 Dec 1994 15:19:56 -0500 (EST) From: Steve Burris Subject: Calculus: Please help me! How do I find the slope of: root x = -ln (xy) at (4,2) The term on the left is supposed to be the square root of x. Thnks a lot... :) Date: Wed, 14 Dec 1994 16:19:16 -0500 (EST) From: Dr. Ken Subject: Re: Calculus: Please help me! Hello there! Well I can't exactly answer your question because the point (4,2) is not on the graph. Instead I will find teh slope of the tangent line when x=4 and I will let you figure out what the proper y value is. There are two ways to approch your problem. Probably the easiest way to do it is to solve for y, and then take the derivative with respect to x. So if we raise e to the power of each side of the equation, we get e^(-Sqrt{x}) = e^(Log(xy)) = xy And then we divide through by x to get y = (x*e^Sqrt{x})^-1 So now we take the derivative of both sides, and we find that dy/dx = -(x*e^Sqrt{x})^-2 * (1*e^Sqrt{x} + x*e^Sqrt{x}/2Sqrt{x}) Do you see how that works? If you have trouble with this step, write back with a specific question and we'll explain what we did. Then to get the answer, you plug in 4 for x, and simplify. -(e^2 + 4*e^2 /4) -1 dy/dx at 4 = ----------------- = ------- 4^2*e^4 8 e^2 So that's the answer as I see it. The other way to do the problem is to use implicit differentiation, which is easier when you know how to do it, but kind of tough to learn. Thanks for the question! -Ken "Dr." Math |
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