How do I handle square roots in partial derivatives?

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In summary, the conversation discusses the process of taking partial derivatives, specifically when dealing with square roots. The participants suggest using implicit differentiation and the chain rule to simplify the process. They also provide an example and a step-by-step explanation of how to find the partial derivative of a function with a square root. The conversation concludes with the individual expressing gratitude for the help and realizing that they have forgotten a basic operation.
  • #1
Arden1528
All right, I know that partial derv. are not all that hard. All you do in let's 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.
 
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  • #2
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)

Please forgive the sloppy notation.
 
  • #3
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
 
  • #4
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)"

Just where is your difficulty?
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
 
  • #5
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
 

Related to How do I handle square roots in partial derivatives?

What is a partial derivative?

A partial derivative is a mathematical concept used to find the rate of change of a multivariable function with respect to one of its variables while holding all other variables constant. It is denoted by ∂ (pronounced "del").

Why do we need to handle square roots in partial derivatives?

Sometimes, the function we are taking the partial derivative of may contain a square root. In order to find the derivative, we need to handle the square root appropriately using specific rules and techniques.

What is the rule for handling square roots in partial derivatives?

The general rule for handling square roots in partial derivatives is to rewrite the function using fractional exponents and then apply the power rule. For example, √x can be rewritten as x^(1/2) and the derivative would be (1/2)x^(-1/2).

Can we handle square roots in partial derivatives without rewriting the function?

Yes, there are specific cases where we can handle square roots without rewriting the function. For example, if the function is in the form √(ax + b), we can use the chain rule and the power rule to find the derivative without rewriting the function.

Are there any other techniques for handling square roots in partial derivatives?

Yes, there are other techniques such as using logarithmic differentiation or implicit differentiation. These techniques may be useful for more complex functions containing square roots.

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