Hello,
i have a problem
x^2y^3+2y=6 solve using implicit differentiation.
so am i suppose to take the derivative first?
which would be 3x^2y^2+2xy^3-2=0?
We are given the implicit relationship:
$\displaystyle x^2y^3+2y=6$
Now, using the product and chain rules, we find:
$\displaystyle x^2\left(3y^2\frac{dy}{dx} \right)+2x(y^3)+2\frac{dy}{dx}=0$
Now, solve for $\displaystyle \frac{dy}{dx}$
That's incorrect...in general you want to arrange with the terms having dy/dx as a factor on one side and the other terms on the other. Then factor out dy/dx, and divide through by the other factor to isolate dy/dx. Give this another try...
Let's go back to:
$\displaystyle x^2\left(3y^2\frac{dy}{dx} \right)+2x(y^3)+2\frac{dy}{dx}=0$
Clean this up some:
$\displaystyle 3x^2y^2\frac{dy}{dx}+2xy^3+2\frac{dy}{dx}=0$
Arrange with dy/dx terms on the left, all else on the right:
$\displaystyle 3x^2y^2\frac{dy}{dx}+2\frac{dy}{dx}=-2xy^3$
Factor out dy/dx on the left side:
$\displaystyle \frac{dy}{dx}(3x^2y^2+2)=-2xy^3$
Divide through by $\displaystyle 3x^2y^2+2$
$\displaystyle \frac{dy}{dx}=-\frac{2xy^3}{3x^2y^2+2}$
Just in case a picture helps (and otherwise to correct a small slip/typo, in Mark's differentiation of y^3)...
... where (key in spoiler) ...
Spoiler:
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
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