Implicit DifferentiationDate: 11/03/98 at 00:44:37 From: Roger Mong Subject: Implicit Differentiation I am a student who is very interested in calculus. I have some knowledge of how to differentiate functions in the form y = f(x). However, how do you differentiate an equation that cannot be expressed in terms of x or y? i.e. x + x^y = y^x I know this has to do with Implicit Differentiation, but I don't know how. Can you please help me? Date: 11/03/98 at 07:41:07 From: Doctor Jerry Subject: Re: Implicit Differentiation Hi Roger, Assuming that the equation x + x^y = y^x can be solved for y in terms of x (this might require assumptions about allowable values of x), one thinks about y as a function of x and applies the chain rule. For example, if y is a function of x, then the derivative of x*y with respect to x is 1*y+x*y'. I just applied the product rule. Since I don't have y as an explicit function of x, I can only write y' for the derivative of y with respect to x. If x^2 + y^2 = 1, for example, (which can be solved for y in terms of x, namely, y=sqrt(1-x^2), or -sqrt(1-x^2), we can differentiate as: 2x + 2y*y' = 0 Solve for y': y' = -x/y. This may be as far as you can go. You can calculate the slope at any point (x,y) of the graph but, unless you know y as a function of x, you can't find y' as an explicit function of x. Note that in this case we know y=sqrt(1-x^2) and so y' = -2x/(2*sqrt(1 - x^2)) = -x/y For x + x^y = y^x, we have x + e^(y*ln(x)) = e^(x*ln(y)) Differentiating, 1 + e^(y*ln(x))(y'*ln(x) + y/x) = e^(x*ln(y))(ln(y) + (x/y)y') 1 + x^y(y'*ln(x) + y/x) = y^x(ln(x) + x*y'/y) You can now solve this expression for y' in terms of x and y. As you can see, the entire procedure rests on the chain rule. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/