How do you find the second derivative for v = (xy)/(x-y)

Jase Rocha 2022-09-25 Answered
How do you find the second derivative for v = x y x - y
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Answers (1)

Kaiden Stevens
Answered 2022-09-26 Author has 12 answers
I'm don't know what you're looking for. The function v = x y x - y is a function of two variables. It doesn't have one second derivative, it has 3 second partial derivatives.
Did you like this example?
Subscribe for all access

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

You might be interested in

asked 2022-09-27
I'm given x 3 + y 3 = 6 x y. It's stated that y is a function of x and I'm tasked to differentiate with respect to x.The implicit differentiation is:
3 x 2 + 3 y 2 y = 6 x y + 6 y
Now simplify and express in terms of y . I'm going to number these steps.
1. x 2 + y 2 y = 2 x y + 2 y
2. y = 2 x y + 2 y x 2 y 2
3. y = 2 x y y 2 + 2 y x 2 y 2
4. y = y 2 x y 2 + 2 y x 2 y 2
5. y y = 2 x y 2 + 2 y x 2 y 2 . We know that y y = 1 and therefore I've made a mistake in my algebra. But I'm not sure what is wrong.Here is the correct simplification, taking it from step 1 again:
1. x 2 + y 2 y = 2 x y + 2 y
2. y 2 y 2 x y = 2 y x 2
3. y ( y 2 2 x ) = 2 y x 2
4. y = 2 y x 2 y 2 2 x
This makes complete sense. What was wrong with my simplification? If it's not wrong, how can I arrive at the correct final expression of y ?
asked 2022-09-24
I have the following expression which I need to implicitly differentiate:
x y 2 + x 2 + y + sin ( x 2 y ) = 0
I'm a little confused as I'm not entirely sure what to do with the trig function. Here is my work so far:
d y d x [ x y 2 + x 2 + y + sin ( x 2 y ) ] = d y d x 0
d y 2 d x + 2 x + d y d x + cos ( x 2 y ) ( 2 x d y d x ) = 0
How should I proceed?
asked 2022-08-19
Let z = z ( x , y ) be defined implicitly by F ( x , y , z ( x , y ) ) = 0, where F is a given function of three variables.
Prove that if z ( x , y ) and F are differentiable, then
d z d x = d F d x d F d z
if d F / d z 0
I am kind of confused with z = z ( x , y ). Should I differentiate d z/ d x and d z/ d y, but there's nothing inside x and y so what should I differentiate d x and d y with then? I've been told that this is something related to implicit differentiation. In some other examples, they do use differentiate F i.e. d F but in most examples, I only see that they are differentiating the variables inside it.
asked 2022-07-17
How do you implicitly differentiate x 2 + 2 x y - y 2 + x = 2
asked 2022-11-20
Please excuse if the formatting of this post is wrong.

There's a question that asks for the 2nd derivative of y 2 x 3 x y = 2

From what I know, I have to use implicit differentiation, using which I get:
12 + 18 y ( 1 3 x ) 2
But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)? By doing that I got:
48 ( 1 3 x ) 3
I'm not sure if this is a correct answer as I am new to differentiation. I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?
asked 2022-08-10
As we know, for implicit differentiation we have y x = F x F y
but if I use a trick that y x = F x F y = F x y F = y x as F cancels each other out, so it is like y x = F x F y without the "minus", where did I go wrong?
asked 2022-08-19
I would like to find d y d x , where we have the implicit equation
( 2 x + y ) 4 + 3 x 2 y = y 3 .
I just can't seem to understand the problem and which rule of calculus to use. Could someone please further elaborate?

New questions