# Math Help - dy/dx

1. ## dy/dx

I got a question:
Find dy/dx of:

(x^3)+(y^3)-(2x)=3

Now what i got is:
((-3x^2)-2) / (3y^2)

By doing this:
(3x^2)+(3y^2)dy/dx-2=0

what am i doing wrong?

2. It looks right, I think you just messed up the negative signs at the end a bit.

$x^3+y^3-2x=3$

$3x^2+3y^2\frac{dy}{dx}-2=0$

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

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

3. and this one:

((x^2)/4) + ((y^2)/16) = 1

i did this:
((8x-(x^2))/16) + ((32y-(y^2))/256 dy/dx

Now from here am confused if its right?

4. Originally Posted by taurus
((x^2)/4) + ((y^2)/16) = 1
To derive this, you gotta use the Implicit Differentiation.

$\frac14x^2+\frac1{16}y^2=1$

Since we want $\frac{dy}{dx}$, we use the chain rule for $y$, this yields

$\frac12x+\frac18yy'=0$

Now just make $y'$ the subject of the equation.

5. that doesnt yield the correct answer? could you run through the steps pls