dy dx of the curve

Apr 2009
136
12
find dydx of

x^2 + 3xy + 2y^2 = 5

My working

2y^2 = 5 - x^2 - 3xy

y^2 = 1/2 . (5 - x^2 - 3xy)

y = root 1/2 . (5 - x^2 - 3xy)

y = 1/2 . (5 - x^2 - 3xy)^1/2

y = (2.5 - 1/2x^2 - 1.5xy)^1/2

y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

Advice please if i am close
Cheers
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
find dydx of

x^2 + 3xy + 2y^2 = 5

My working

2y^2 = 5 - x^2 - 3xy

y^2 = 1/2 . (5 - x^2 - 3xy)

y = root 1/2 . (5 - x^2 - 3xy)

y = 1/2 . (5 - x^2 - 3xy)^1/2

y = (2.5 - 1/2x^2 - 1.5xy)^1/2

y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

Advice please if i am close
Cheers
first of all, it's dy/dx , not dydx.

second, you're not even close in your "method".

research finding implicit derivatives in your text or on the web.
 
Jan 2010
354
173
Just to clarify a bit on what skeeter is saying:

To use the more "traditional" method of solving for a derivative, you would need to solve for y explicitly. What that means is that you can get the equation into the form \(\displaystyle y=\cdots\) where there are no \(\displaystyle y\) terms on the right hand side at all.

Although your work to solve for \(\displaystyle y\) seemed like it was on the right track, the problem is you are stuck with a \(\displaystyle y\) term in the right hand side.

The correct way to handle this problem is to use implicit differentiation. This method is used without first solving for \(\displaystyle y\).

Just to give you a small hint -- You do not need to simply or modify the equation at all before taking the derivative! Have you learned implicit differentiation techniques?
 
Oct 2009
35
11
find dydx of

x^2 + 3xy + 2y^2 = 5

My working

2y^2 = 5 - x^2 - 3xy

y^2 = 1/2 . (5 - x^2 - 3xy)

y = root 1/2 . (5 - x^2 - 3xy)

y = 1/2 . (5 - x^2 - 3xy)^1/2

y = (2.5 - 1/2x^2 - 1.5xy)^1/2

y1 = 1/2 (2.5 - 1/2x^2 - 1.5xy)^-1/2. 3

Advice please if i am close
Cheers
When you have \(\displaystyle f(y)\), it is the same as evaluating \(\displaystyle f(x)\) except that there is a \(\displaystyle \frac{dy}{dx}\) attached (multiplied).

So deriving \(\displaystyle y^2 = x\) gives

\(\displaystyle 2y\frac{dy}{dx} = 1\)

\(\displaystyle \frac{dy}{dx} = \frac{1}{2y}\)

This is simply a special case of the chain rule: notice that when evaluating x's this rule is hidden by the fact \(\displaystyle \frac{dx}{dx} = 1\).

Product rules with \(\displaystyle f(y)g(x)\) work as you would expect. Can you use this technique to answer your question?