How do you find the second derivative for $v=\frac{xy}{x-y}$

Jase Rocha
2022-09-25
Answered

How do you find the second derivative for $v=\frac{xy}{x-y}$

You can still ask an expert for help

Kaiden Stevens

Answered 2022-09-26
Author has **12** answers

I'm don't know what you're looking for. The function $v=\frac{xy}{x-y}$ is a function of two variables. It doesn't have one second derivative, it has 3 second partial derivatives.

asked 2022-09-27

I'm given ${x}^{3}+{y}^{3}=6xy$. It's stated that $y$ is a function of $x$ and I'm tasked to differentiate with respect to $x$.The implicit differentiation is:

$3{x}^{2}+3{y}^{2}{y}^{\prime}=6x{y}^{\prime}+6y$

Now simplify and express in terms of ${y}^{\prime}$. I'm going to number these steps.

1. ${x}^{2}+{y}^{2}{y}^{\prime}=2x{y}^{\prime}+2y$

2. ${y}^{\prime}=\frac{2x{y}^{\prime}+2y-{x}^{2}}{{y}^{2}}$

3. ${y}^{\prime}=\frac{2x{y}^{\prime}}{{y}^{2}}+\frac{2y-{x}^{2}}{{y}^{2}}$

4. ${y}^{\prime}={y}^{\prime}\ast \frac{2x}{{y}^{2}}+\frac{2y-{x}^{2}}{{y}^{2}}$

5. $\frac{{y}^{\prime}}{{y}^{\prime}}=\frac{2x}{{y}^{2}}+\frac{2y-{x}^{2}}{{y}^{2}}.$ We know that $\frac{{y}^{\prime}}{{y}^{\prime}}=1$ 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. ${x}^{2}+{y}^{2}{y}^{\prime}=2x{y}^{\prime}+2y$

2. ${y}^{2}{y}^{\prime}-2x{y}^{\prime}=2y-{x}^{2}$

3. ${y}^{\prime}({y}^{2}-2x)=2y-{x}^{2}$

4. ${y}^{\prime}=\frac{2y-{x}^{2}}{{y}^{2}-2x}$

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 ${y}^{\prime}$?

$3{x}^{2}+3{y}^{2}{y}^{\prime}=6x{y}^{\prime}+6y$

Now simplify and express in terms of ${y}^{\prime}$. I'm going to number these steps.

1. ${x}^{2}+{y}^{2}{y}^{\prime}=2x{y}^{\prime}+2y$

2. ${y}^{\prime}=\frac{2x{y}^{\prime}+2y-{x}^{2}}{{y}^{2}}$

3. ${y}^{\prime}=\frac{2x{y}^{\prime}}{{y}^{2}}+\frac{2y-{x}^{2}}{{y}^{2}}$

4. ${y}^{\prime}={y}^{\prime}\ast \frac{2x}{{y}^{2}}+\frac{2y-{x}^{2}}{{y}^{2}}$

5. $\frac{{y}^{\prime}}{{y}^{\prime}}=\frac{2x}{{y}^{2}}+\frac{2y-{x}^{2}}{{y}^{2}}.$ We know that $\frac{{y}^{\prime}}{{y}^{\prime}}=1$ 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. ${x}^{2}+{y}^{2}{y}^{\prime}=2x{y}^{\prime}+2y$

2. ${y}^{2}{y}^{\prime}-2x{y}^{\prime}=2y-{x}^{2}$

3. ${y}^{\prime}({y}^{2}-2x)=2y-{x}^{2}$

4. ${y}^{\prime}=\frac{2y-{x}^{2}}{{y}^{2}-2x}$

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 ${y}^{\prime}$?

asked 2022-09-24

I have the following expression which I need to implicitly differentiate:

$x{y}^{2}+{x}^{2}+y+\mathrm{sin}({x}^{2}y)=0$

I'm a little confused as I'm not entirely sure what to do with the trig function. Here is my work so far:

$\frac{dy}{dx}[x{y}^{2}+{x}^{2}+y+\mathrm{sin}({x}^{2}y)]=\frac{dy}{dx}0$

$\frac{d{y}^{2}}{dx}+2x+\frac{dy}{dx}+\mathrm{cos}({x}^{2}y)(2x\frac{dy}{dx})=0$

How should I proceed?

$x{y}^{2}+{x}^{2}+y+\mathrm{sin}({x}^{2}y)=0$

I'm a little confused as I'm not entirely sure what to do with the trig function. Here is my work so far:

$\frac{dy}{dx}[x{y}^{2}+{x}^{2}+y+\mathrm{sin}({x}^{2}y)]=\frac{dy}{dx}0$

$\frac{d{y}^{2}}{dx}+2x+\frac{dy}{dx}+\mathrm{cos}({x}^{2}y)(2x\frac{dy}{dx})=0$

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

$\frac{dz}{dx}=-\frac{\frac{dF}{dx}}{\frac{dF}{dz}}$

if $dF/dz\ne 0$

I am kind of confused with $z=z(x,y)$. Should I differentiate $dz$/$dx$ and $dz$/$dy$, but there's nothing inside $x$ and $y$ so what should I differentiate $dx$ and $dy$ 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. $dF$ but in most examples, I only see that they are differentiating the variables inside it.

Prove that if $z(x,y)$ and $F$ are differentiable, then

$\frac{dz}{dx}=-\frac{\frac{dF}{dx}}{\frac{dF}{dz}}$

if $dF/dz\ne 0$

I am kind of confused with $z=z(x,y)$. Should I differentiate $dz$/$dx$ and $dz$/$dy$, but there's nothing inside $x$ and $y$ so what should I differentiate $dx$ and $dy$ 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. $dF$ but in most examples, I only see that they are differentiating the variables inside it.

asked 2022-07-17

How do you implicitly differentiate ${x}^{2}+2xy-{y}^{2}+x=2$

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 $y-2x-3xy=2$

From what I know, I have to use implicit differentiation, using which I get:

$\frac{12+18y}{(1-3x{)}^{2}}$

But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)? By doing that I got:

$\frac{48}{(1-3x{)}^{3}}$

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?

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

From what I know, I have to use implicit differentiation, using which I get:

$\frac{12+18y}{(1-3x{)}^{2}}$

But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)? By doing that I got:

$\frac{48}{(1-3x{)}^{3}}$

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 $\frac{\mathrm{\partial}y}{\mathrm{\partial}x}=-\frac{{F}_{x}}{{F}_{y}}$

but if I use a trick that $\frac{\mathrm{\partial}y}{\mathrm{\partial}x}=\frac{\frac{\mathrm{\partial}F}{\mathrm{\partial}x}}{\frac{\mathrm{\partial}F}{\mathrm{\partial}y}}=\frac{\mathrm{\partial}F}{\mathrm{\partial}x}\frac{\mathrm{\partial}y}{\mathrm{\partial}F}=\frac{\mathrm{\partial}y}{\mathrm{\partial}x}$ as $\mathrm{\partial}F$ cancels each other out, so it is like $\frac{\mathrm{\partial}y}{\mathrm{\partial}x}=\frac{{F}_{x}}{{F}_{y}}$ without the "minus", where did I go wrong?

but if I use a trick that $\frac{\mathrm{\partial}y}{\mathrm{\partial}x}=\frac{\frac{\mathrm{\partial}F}{\mathrm{\partial}x}}{\frac{\mathrm{\partial}F}{\mathrm{\partial}y}}=\frac{\mathrm{\partial}F}{\mathrm{\partial}x}\frac{\mathrm{\partial}y}{\mathrm{\partial}F}=\frac{\mathrm{\partial}y}{\mathrm{\partial}x}$ as $\mathrm{\partial}F$ cancels each other out, so it is like $\frac{\mathrm{\partial}y}{\mathrm{\partial}x}=\frac{{F}_{x}}{{F}_{y}}$ without the "minus", where did I go wrong?

asked 2022-08-19

I would like to find $\frac{\mathrm{d}y}{\mathrm{d}x}$, where we have the implicit equation

$(2x+y{)}^{4}+3{x}^{2}y={y}^{3}.$

I just can't seem to understand the problem and which rule of calculus to use. Could someone please further elaborate?

$(2x+y{)}^{4}+3{x}^{2}y={y}^{3}.$

I just can't seem to understand the problem and which rule of calculus to use. Could someone please further elaborate?