# Partial Derv.

Arden1528
All right, I know that partial derv. are not all that hard. All you do in lets say f(x)= x^2+2yx+3y is find the dervitives of x while you treat y as a constent, and vice versa. But I keep running into problems having square roots. I hat these things.
One example could be f(x)= sqr(20-x^2-7y^2). I know yo treat the y as a constent, but I am horrible with my derv. Any help would be appreciated.

If you hate square roots so much, would you be more comfortable with implicit differentiation? Can you easily take the deriviatve of g(x)=20-x2-7y2? Sure you can. Now, g(x)=[f(x)]2 and 2ff'=g' so f'=g'/(2f) which agrees with what you know about square roots. You know f, so all you need is g', which is easy to find.
g'=-2x
So the partial derivative of f wrt x is,
f'=(-2x)/(2f(x))=-x/f(x)

Hurkyl
Staff Emeritus
Gold Member
The other way is to apply the chain rule to the power rule:

(d/dx) (u^n) = n u^(n-1) (du/dx)

then just plug in the right things for u and n and finish the rest of the work

Soroban
Hello, Arden1528!

You said, "I keep running into problems having square roots.
One example could be f(x,y)= sqr(20 - x^2 - 7y^2)"

You know how to take partial derivatives.
You know that a square root is a one-half power.
And you know the Chain Rule.

f(x,y) = (20 - x2 - 7y2)1/2

Hence, fx = (1/2)(20 - x2 - 7y2)-1/2(-2x) = -x/(20 - x2 - 7y2)1/2

And, fy = {1/2)(20 - x2 - 7y2)-1/2(-14y) = -7y/(20 - x2 - 7y2)1/2

Arden1528
I did not even think about just putting the problems to the (1/2) power. I seem to forget the things that would make these problems easier. But thanks for all the help. I feel like I just forgot a really elementry operation...thanks