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### Implicit Differentiation

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Date: 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
```

```
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/
```
Associated Topics:
High School Calculus

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