Question

dy/dx of xy^2+4y-5x^2
(implicit differentation) plz explain how to do this problem step by step...plz and ty..

Answer 1

We differentiate with respect to 'x' if for one term there is a 'y' we differentiate it and multiply it by dy/dx or y'.
So for your question:
Using implicit differentiation:
Differentiate xy^2 using the chain rule: y^2+2xy*(dy/dx). Do you understand this?

Answer 2

yes but i would appreciate if u could work out the porblem so i can use it as an example on similar problems like this that could be on my exam tmmrw and i dnt fail that exam..

Answer 3

\[xy^2+4y-5x^2=0\] Is this really your equation?

Answer 4

Differentiate xy^2:
Using chain rule: differentiate with respect to x and treat the y term as a constant (y^2) then differentiate with respect to y and treat the x term as a constant (2xy*dy/dx). Ans.: y^2+2xy*dy/dx.
Differentiate 4y:
Because we are asked to differentiate with respect to x if we differentiate with respect to y we multiply our answer by dy/dx always. Ans.: 4*dy/dx.
Differentiate 5x^2 = 10x.

Complete answer: y^2 + 2xy*dy/dx+4*dy/dx -10x.

Answer 5

I always look at "y" as though it's derivative is "dy/dx". This helps when looking at the chain rule when taking the derivative of the outside times the derivative of the inside.

For example, the derivative of:

3y^5

is:

15y^5*(dy/dx)

Does that explain it?

Then what you do is solve for (dy/dx) after you've differentiated with algebra similar to how you would normally solve for something like x.

Answer 6

its*

Answer 7

derivative of 3y^5 is 15y^4 dy/dx

Answer 8

lol my bad, but yes you get the point?

Answer 9

\[(uv)'=uv'+u'v\]

Answer 10

That is the product rule

Answer 11

http://openstudy.com/study#/updates/50d4035ae4b0d6c1d54146dc

Answer 12

yup u are right it is the product rule not the chain rule. @ChmE

Answer 13

But you were also right in saying that it is also the chain rule. Both are used

Answer 14

Implicit Differentiation:

Hmm, how should I explain it.

Basically it is showing the chain rule instead of it just staying hidden.

The derivative of y^2 is 2y. I think we all would agree on that but what don't we see?

It is really y^2 * y' and since y' = 1 we just leave that out.

But if wanted to take the derivative of (2x+1)^2 it would be 2(2x+1) * 2 because the derivative of y (in this case y=2x + 1) is equal to 2, so it is not hidden here.

All implicit differentiation really is, is showing the chain rule as we differentiate. y' is the same as dy/dx. dy/dx means that we are taking the derivative of y with respect to x. So after every derivative of y we have to include dy/dx after it.

The only explanation for why we don't do dx/dy after the x is because we are only trying to solve for dy/dx. If anybody has a better explanation for this part plz feel free to add.

After we differentiate we need to solve for dy/dx. This is just all algebra.

Ex)
3x^2 + 2y^3 = 0
6x + 6y^2 dy/dx = 0

Now we have to solve for dy/dx though

6y^2 dy/dx = -6x
dy/dx = -6x/6y^2
dy/dx = -x/y^2

Answer 15

\[x \color{blue}{y^2}+4 \color{blue}{y}-5x^2=0\]\[2x \color{blue}{y} \color{red}{\frac{dy}{dx}}+\color{blue}{y^2}+4 \color{red}{\frac{dy}{dx}}-10x=0\]\[\color{red}{\frac{dy}{dx}}(2x \color{blue}{y}+4)=10x-\color{blue}{y^2}\]\[\color{red}{\frac{dy}{dx}}=\frac{10x-\color{blue}{y^2}}{2x \color{blue}{y}+4}\]

Answer 16

lolol nice :)

Answer 17

Oh it's on now. Colors are the best!!

Answer 18

"The only explanation for why we don't do dx/dy after the x is because we are only trying to solve for dy/dx."

Wouldn't it make more sense to say, we don't use dx/dy because we're differentiating the equation with respect to x? dx/dy would mean we're differentiating the equation with respect to y.

Answer 19

Nice I like that