derivative very tricky : need guide

**rcs** · July 30th 2012, 11:45 PM

How are these pattered in the rule of derivatives?
Find the derivative of dy/dx of the relations:

1.) x^3y+ 2y^4 - x^4 = 2xy

and

2.) x^2 = x+2y/ x - 2y

thanks a lot

**jgv115** · July 31st 2012, 12:45 AM

If I want to simplify $\frac{d(y)}{dx}$ , I obviously can't do it that easily as the variables are wrong.

However, I can apply the chain rule: $\frac{d(y)}{dx}.\frac{dx}{dy}=\frac{d(y)}{dy}$

Solving for $\frac{d(y)}{dx}$ , I get $\frac{d(y)}{dx}=\frac{d(y)}{dy}.\frac{dy}{dx}$

Evaluate $\frac{d(y)}{dy}$ which is $1$ . Therefore $\frac{d(y)}{dx}= \frac{dy}{dx}$

Now try apply these principles in your question

**rcs** · July 31st 2012, 01:35 AM

ill try sir... ill be back and let you see my answer also... hope i can get it right

thanks

**emakarov** · July 31st 2012, 01:40 AM

Originally Posted by jgv115

If I want to simplify $\frac{d(y)}{dx}$ , I obviously can't do it that easily as the variables are wrong.

However, I can apply the chain rule: $\frac{d(y)}{dx}.\frac{dx}{dy}=\frac{d(y)}{dy}$

Solving for $\frac{d(y)}{dx}$ , I get $\frac{d(y)}{dx}=\frac{d(y)}{dy}.\frac{dy}{dx}$

Evaluate $\frac{d(y)}{dy}$ which is $1$ . Therefore $\frac{d(y)}{dx}= \frac{dy}{dx}$

Now try apply these principles in your question

To be honest, I don't understand this. What is $\frac{d(y)}{dx}$ and how is it different from $\frac{dy}{dx}$ ?

OP, it seems that you need to differentiate an implicit function. Obviously, you need to read how to do this, for example, in Wikipedia. Once you learn the method and read several finished examples, please describe the exact difficulty you are having with this particular problem. The thing is that this forum is not a tutorial service and you can't expect people to explain completely new material to you. The forum is here to help those who know the material but struggle to apply it in a concrete situation.

**rcs** · July 31st 2012, 03:36 AM

thank you

**Deveno** · July 31st 2012, 03:37 AM

i'll give you an example, perhaps you'll see how it works:

if x²+y² = 1, find dy/dx.

step 1) differentiate both sides with respect to x:

2x(dx/dx) + 2y(dy/dx) = 0 but dx/dx = 1, so:

2x + 2y(dy/dx) = 0

step 2) solve for dy/dx:

2y(dy/dx) = -2x

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

**x3bnm** · July 31st 2012, 04:40 PM

So this is called implicit differentiation. You can differentiate both sides with respect to $x$ where $y$ is a function of $x$ and $\frac{dx}{dx} = 1$ :

$\begin{aligned} x^3 y + 2 y^4 -x^4 & = 2xy \\ x^3 \frac{dy}{dx} + 3x^2 y + 8y^3 \frac{dy}{dx} - 4x^3 & = 2y + 2x \frac{dy}{dx} \\ \\ [\text{ Now you gather dy/dx one side and isolating dy/dx }] \\ \\ \frac{dy}{dx} & = \frac{2y - 3x^2 y + 4 x^3}{x^3 + 8 y^3 - 2x} \end{aligned}$

For the second one:

$\begin{aligned} x^2 & = x + \frac{2y}{x} - 2y \\ 2x & = 1 + \frac{x \frac{2 dy}{dx} - 2y}{x^2} - 2 \frac{dy}{dx} \\ \\ [\text{Same as before gather dy/dx together and simplifying using Algebra you get }] \\ \\ \frac{dy}{dx} & = \frac{2x^3 -x^2 + 2y}{2x - 2x^2} \end{array}$

You can check it to confirm here for the first equation:

differentiation x^3 y +2 y^4 -x^4 - 2xy = 0 - Wolfram|Alpha

And for the second one:

differentiation x^2 - x - 2y/x +2y = 0 - Wolfram|Alpha

Hope this helps.

**rcs** · July 31st 2012, 05:28 PM

a million of thanks for the solution sir... the best Helper of this site... --- rcs ----

**x3bnm** · July 31st 2012, 06:11 PM

Originally Posted by rcs

a million of thanks for the solution sir... the best Helper of this site... --- rcs ----

You're welcome. Glad to help.

**rcs** · September 2nd 2012, 06:32 PM

Originally Posted by x3bnm

So this is called implicit differentiation. You can differentiate both sides with respect to $x$ where $y$ is a function of $x$ and $\frac{dx}{dx} = 1$ :

$\begin{aligned} x^3 y + 2 y^4 -x^4 & = 2xy \\ x^3 \frac{dy}{dx} + 3x^2 y + 8y^3 \frac{dy}{dx} - 4x^3 & = 2y + 2x \frac{dy}{dx} \\ \\ [\text{ Now you gather dy/dx one side and isolating dy/dx }] \\ \\ \frac{dy}{dx} & = \frac{2y - 3x^2 y + 4 x^3}{x^3 + 8 y^3 - 2x} \end{aligned}$

$\begin{aligned} x^2 & = x + \frac{2y}{x} - 2y \\ 2x & = 1 + \frac{x \frac{2 dy}{dx} - 2y}{x^2} - 2 \frac{dy}{dx} \\ \\ [\text{Same as before gather dy/dx together and simplifying using Algebra you get }] \\ \\ \frac{dy}{dx} & = \frac{2x^3 -x^2 + 2y}{2x - 2x^2} \end{array}$

You can check it to confirm here for the first equation:

differentiation x^3 y +2 y^4 -x^4 - 2xy = 0 - Wolfram|Alpha

And for the second one:

differentiation x^2 - x - 2y/x +2y = 0 - Wolfram|Alpha

Hope this helps.

i found it confusing .. how did you get this
For the second one:

x^2 = 1 + 2y/x - 2y sir how did you this ?

because it is supposed to be in number 2: the equation is x^2 = (x+2y)/(x-2y)

thanks

**Prove It** · September 2nd 2012, 08:05 PM

x3bnm read the question AS YOU WROTE IT. If you had used the brackets in the correct place in the first place, you would have been given a correct solution.

Anyway, you now know how to differentiate implicitly, so I suggest you try it out for yourself with the correct question.

**rcs** · September 3rd 2012, 05:05 AM

Originally Posted by Prove It

x3bnm read the question AS YOU WROTE IT. If you had used the brackets in the correct place in the first place, you would have been given a correct solution.

Anyway, you now know how to differentiate implicitly, so I suggest you try it out for yourself with the correct question.

is it possible to get the derivative like ,
x^2(x-2y) = x + 2y sir?
multiplying both side by x - 2y of the equation, before taking the derivative

**x3bnm** · September 4th 2012, 03:00 PM

Originally Posted by rcs

is it possible to get the derivative like ,
x^2(x-2y) = x + 2y sir?
multiplying both side by x - 2y of the equation, before taking the derivative

Yes the new solution is given below:

$\begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\ 2x =& \frac{(x - 2y)(1 + 2 \frac{dy}{dx}) - (x+2y)(1 - 2 \frac{dy}{dx})}{(x-2y)^2} \\ 2x =& \frac{x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx}}{(x -2y)^2} \\ 2x(x - 2y)^2 =& x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx} \\ 2x(x - 2y)^2 + 4y =& 4x \frac{dy}{dx} \\ \frac{dy}{dx} =& \frac{2x(x - 2y)^2 + 4y}{4x} \\ \frac{dy}{dx} =& \frac{x(x-2y)^2 + 2y}{2x} \\ \frac{dy}{dx} =& \frac{x^3 - 4x^2 y + 4 xy^2 + 2y}{2x} \text{.................[after simplification] } \end{align*}$

Hope it will help you.

Also you can check the answer at:

differentiation x^2 = (x+2y)/(x-2y) - Wolfram|Alpha

**rcs** · September 7th 2012, 05:50 AM

Originally Posted by x3bnm

Yes the new solution is given below:

$\begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\ 2x =& \frac{(x - 2y)(1 + 2 \frac{dy}{dx}) - (x+2y)(1 - 2 \frac{dy}{dx})}{(x-2y)^2} \\ 2x =& \frac{x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx}}{(x -2y)^2} \\ 2x(x - 2y)^2 =& x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx} \\ 2x(x - 2y)^2 + 4y =& 4x \frac{dy}{dx} \\ \frac{dy}{dx} =& \frac{2x(x - 2y)^2 + 4y}{4x} \\ \frac{dy}{dx} =& \frac{x(x-2y)^2 + 2y}{2x} \\ \frac{dy}{dx} =& \frac{x^3 - 4x^2 y + 4 xy^2 + 2y}{2x} \text{.................[after simplification] } \end{align*}$

Hope it will help you.

Also you can check the answer at:

differentiation x^2 = (x+2y)/(x-2y) - Wolfram|Alpha

thanks x3bmn you are also nice and kind. God BLess

**rcs** · September 7th 2012, 06:02 AM

Such a big help sir. Thanks

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