# Implicit Differentiation

• October 3rd 2009, 04:51 PM
isp_of_doom
Implicit Differentiation
I'm currently studying maths, and need help with the following question:

$x^4 + 4yx - 2y^5 = 3$

I'm new to implicit differentiation, and have followed the book closely on this one (using dy/dx notation, which I am also new to). I've been using matlab to verify my answers, but it hasn't matched with this one.

My answer $dy/dx = \frac{(4x^3 + 4y + 4x)}{-10y^4}$
Matlab answer = $4x^3 + 4y$

• October 3rd 2009, 09:43 PM
Jameson
When you want an error spotted in your work, it's best to show what you did step by step. Otherwise I have no idea where you made a mistake and the only way to help you is to type out a full solution myself, which is time consuming.

Can you post your work?
• October 3rd 2009, 11:10 PM
isp_of_doom
That makes sense... working out step by step below:

Step One
$\frac{d}{dx}(x^4+4xy-2y^5) = \frac{d}{dx}(3)$

Next step: Calculating the derivative bit by bit

$\frac{d}{dx}(x^4) + \frac{d}{dx}(4xy) - \frac{d}{dx}(2y^5) = 0$

Now...finding derivatives (including product rule regarding d/dx(4xy))

$4x^3 + \frac{d}{dx}(4x)y + 4x\frac{d}{dx}(y) - \frac{dy}{dx}\frac{d}{dy}(2y^5) = 0$

this is where I think I get stuck...most likely regarding product rule and dy/dx style notation (what is the difference between $\frac{d}{dx} \frac{dy}{dx} \frac{d}{dy}$ and $\frac{dx}{dy}$?

$4x^3 + 4y + 4x\frac{d}{dx}(y) - \frac{dy}{dx}10y^4 = 0$

$4x^3 + 4y + 4x -\frac{dy}{dx}10x^4 = 0$

$\frac{dy}{dx} = \frac{4x^3 + 4y + 4x}{-10y^4}$

Also, when using dy/dx notation I've noticed my uni-written coursebook uses brackets after dy/dx to denote un-derived components of the equation, then once they've been differentiated the brackets are removed to signify so. Am I right in saying this? (is that the rule of thumb?)