How do you find dy/dx for $y+2x-3x{y}^{3}=4$

Angel Kline
2022-10-14
Answered

How do you find dy/dx for $y+2x-3x{y}^{3}=4$

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aitorarjolia

Answered 2022-10-15
Author has **11** answers

You must remember that y is a function of x so that to take into account this you

have to include $\frac{dy}{dx}$ when deriving any y in your function, as:

$1\frac{dy}{dx}+2-3{y}^{3}-9x{y}^{2}\frac{dy}{dx}=0$

Collecting $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{3{y}^{3}-2}{1-9x{y}^{2}}$

have to include $\frac{dy}{dx}$ when deriving any y in your function, as:

$1\frac{dy}{dx}+2-3{y}^{3}-9x{y}^{2}\frac{dy}{dx}=0$

Collecting $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{3{y}^{3}-2}{1-9x{y}^{2}}$

asked 2022-07-23

I was looking to implicitly differentiate

$-22{x}^{6}+4{x}^{33}y+{y}^{7}=-17$

and found it to be

$\frac{dy}{dx}}={\displaystyle \frac{132{x}^{5}-132{x}^{32}y}{4{x}^{33}+7{y}^{6}}$

Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with

$\frac{0}{11}$

Now I go to solve

$y-y1=m(x-x1)$

getting

$y-1=0(x-1)$

resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?

$-22{x}^{6}+4{x}^{33}y+{y}^{7}=-17$

and found it to be

$\frac{dy}{dx}}={\displaystyle \frac{132{x}^{5}-132{x}^{32}y}{4{x}^{33}+7{y}^{6}}$

Now, I am trying to find the equation of the tangent line to the curve at the coordinate (1,1). So I then plug both 1 in for x and y into the above equation and come up with

$\frac{0}{11}$

Now I go to solve

$y-y1=m(x-x1)$

getting

$y-1=0(x-1)$

resulting in $y=1$ and the equation to be $y=x+1$ for my final answer. Am I going about this in the correct manner?

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 2022-07-23

Why, after differentiating $y$ on one side of the equation, is $dy/dx$ added?

As clarification an example I will provide an example:

Implicitly differentiate ${y}^{2}=x$.

You get

$2y\frac{dy}{dx}=1$

as one of the first steps in differentiation. Why is the $dy/dx$ added after ${y}^{2}$ is differentiated?

As clarification an example I will provide an example:

Implicitly differentiate ${y}^{2}=x$.

You get

$2y\frac{dy}{dx}=1$

as one of the first steps in differentiation. Why is the $dy/dx$ added after ${y}^{2}$ is differentiated?

asked 2022-11-14

I'm supposed to implicitly differentiate the following and give the answer in terms of $y\mathrm{\prime}$

$\mathrm{tan}(x-y)=\frac{y}{1+{x}^{2}}$

$\frac{(1+{x}^{2})y\mathrm{\prime}-2xy}{(1+{x}^{2}{)}^{2}}$

How do I solve for $y\mathrm{\prime}$?

Edit: After a ludicrous amount of algebra, i finally ended up at

$\frac{se{c}^{2}(x-y)(1+{x}^{2}{)}^{2}+2xy}{(1+{x}^{2})(1+se{c}^{2}(x-y)(1+{x}^{2}))}$

which apparently is correct.

$\mathrm{tan}(x-y)=\frac{y}{1+{x}^{2}}$

$\frac{(1+{x}^{2})y\mathrm{\prime}-2xy}{(1+{x}^{2}{)}^{2}}$

How do I solve for $y\mathrm{\prime}$?

Edit: After a ludicrous amount of algebra, i finally ended up at

$\frac{se{c}^{2}(x-y)(1+{x}^{2}{)}^{2}+2xy}{(1+{x}^{2})(1+se{c}^{2}(x-y)(1+{x}^{2}))}$

which apparently is correct.

asked 2022-09-18

$f(x,y)={x}^{2}+{y}^{2}-1$

$0={x}^{2}+{y}^{2}-1\Rightarrow y=\sqrt{1-{x}^{2}}$

I differentiaded $g(x)=\sqrt{1-{x}^{2}}$ with the chain rule and got ${g}^{\prime}(x)=-\frac{x}{\sqrt{1-{x}^{2}}}$.

Can someone tell me how to do it with implicit differentiation?

I tried this formula ${y}^{\prime}(x)=-\frac{{f}_{x}}{{f}_{y}}=-\frac{x}{y}$, the solution should obviously be the same so I guess I might not be allowed to use the formula here? We had the implicit function theorem and $dF(x,y)\left(\begin{array}{c}h\\ k\end{array}\right)=\left(\begin{array}{c}h\\ df(x,y)(h,k)\end{array}\right)$ but I don't really know how to apply this here.

$0={x}^{2}+{y}^{2}-1\Rightarrow y=\sqrt{1-{x}^{2}}$

I differentiaded $g(x)=\sqrt{1-{x}^{2}}$ with the chain rule and got ${g}^{\prime}(x)=-\frac{x}{\sqrt{1-{x}^{2}}}$.

Can someone tell me how to do it with implicit differentiation?

I tried this formula ${y}^{\prime}(x)=-\frac{{f}_{x}}{{f}_{y}}=-\frac{x}{y}$, the solution should obviously be the same so I guess I might not be allowed to use the formula here? We had the implicit function theorem and $dF(x,y)\left(\begin{array}{c}h\\ k\end{array}\right)=\left(\begin{array}{c}h\\ df(x,y)(h,k)\end{array}\right)$ but I don't really know how to apply this here.

asked 2022-08-28

$\mathrm{ln}(1+xy)=xy$

When I try to implicitly differentiate this I get

$\frac{1}{1+xy}(y+x{y}^{\prime})$ = (y + xy')

At which point the $(y+x{y}^{\prime})$ terms cancel out, leaving no ${y}^{\prime}$ to solve for.

However, the answer to this is $-\frac{y}{x}$... How do you get this?

When I try to implicitly differentiate this I get

$\frac{1}{1+xy}(y+x{y}^{\prime})$ = (y + xy')

At which point the $(y+x{y}^{\prime})$ terms cancel out, leaving no ${y}^{\prime}$ to solve for.

However, the answer to this is $-\frac{y}{x}$... How do you get this?

asked 2022-08-12

How do you find the second derivative by implicit differentiation?