How do you differentiate $5xy+{y}^{3}=2x+3y$

braffter92
2022-09-29
Answered

How do you differentiate $5xy+{y}^{3}=2x+3y$

You can still ask an expert for help

recepiamsb

Answered 2022-09-30
Author has **9** answers

You can use implicit differentiation rememberig that y represents a function of x and needs to be differentiated accordingly;

for example: if you have $y}^{2$ you differentiate it to get:

$2y\cdot \frac{dy}{dx}$ where you use $\frac{dy}{dx}$ to take into account the dependence with x.

In your case you get:

$5y+5x\frac{dy}{dx}+3{y}^{2}\frac{dy}{dx}=2+3\frac{dy}{dx}$

collect $\frac{dy}{dx}$

$\frac{dy}{dx}(5x+3{y}^{2}-3)=2-5y$

and

$\frac{dy}{dx}=\frac{2-5y}{5x+3{y}^{2}-3}$

for example: if you have $y}^{2$ you differentiate it to get:

$2y\cdot \frac{dy}{dx}$ where you use $\frac{dy}{dx}$ to take into account the dependence with x.

In your case you get:

$5y+5x\frac{dy}{dx}+3{y}^{2}\frac{dy}{dx}=2+3\frac{dy}{dx}$

collect $\frac{dy}{dx}$

$\frac{dy}{dx}(5x+3{y}^{2}-3)=2-5y$

and

$\frac{dy}{dx}=\frac{2-5y}{5x+3{y}^{2}-3}$

asked 2022-08-26

Given:

${e}^{7z}=xyz$

The task is to compute the partial derivatives dz/dx and dz/dy using implicit differentiation. My solution is now as follows:

$(dz/dx)7{e}^{7z}=yz+(dz/dx)xy$

And so,

$(dz/dx)(7{e}^{7z}-xy)=yz$

Therefore correct answer is,

$\frac{dz}{dx}=\frac{yz}{(7{e}^{7z}-xy)}$

${e}^{7z}=xyz$

The task is to compute the partial derivatives dz/dx and dz/dy using implicit differentiation. My solution is now as follows:

$(dz/dx)7{e}^{7z}=yz+(dz/dx)xy$

And so,

$(dz/dx)(7{e}^{7z}-xy)=yz$

Therefore correct answer is,

$\frac{dz}{dx}=\frac{yz}{(7{e}^{7z}-xy)}$

asked 2022-08-22

I am currently working on a question which involves me differentiating

$\frac{y}{x}$

I can't find nothing in books or on the internet about how to deal with this kind of implicit differentiation.

$\frac{y}{x}$

I can't find nothing in books or on the internet about how to deal with this kind of implicit differentiation.

asked 2022-08-17

If $4{x}^{2}+5x+xy=4$ and $y(4)=-20$, find ${y}^{\prime}(4)$ by implicit differentiation.

I implicitly differentiated the equation, but I don't see how I can use $y(4)=-20$ to my advantage.

I implicitly differentiated the equation, but I don't see how I can use $y(4)=-20$ to my advantage.

asked 2022-07-28

Differentiate the function.

$g(x)=3{x}^{-3}({x}^{4}-3{x}^{3}+15x-4)$

$g(x)=3{x}^{-3}({x}^{4}-3{x}^{3}+15x-4)$

asked 2022-08-11

How do you differentiate ${e}^{xy}+{x}^{2}-{y}^{2}=0$

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 2022-08-14

What is the derivative of $y=6xy$