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?

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- Sep 17th 2007, 06:16 AM #1

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- Sep 17th 2007, 06:25 AM #2

- Sep 17th 2007, 06:33 AM #3

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- Sep 17th 2007, 07:19 AM #4
To derive this, you gotta use the Implicit Differentiation.

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

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

$\displaystyle \frac12x+\frac18yy'=0$

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

- Sep 17th 2007, 07:44 AM #5

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