The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

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
Associated Topics:
High School Calculus

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.