Find y" in terms of x and y by implicit differentiation (x^3)+(y^3)=4^3
So for this is what I have the first derivative will be (3x^2)+ (3y^2)(dy/dx). I know that before I take the second derivative I need to solve for y'. But, I don't know how to do this Please Help.

Question

Find y" in terms of x and y by implicit differentiation (x^3)+(y^3)=4^3
So for this is what I have the first derivative will be (3x^2)+ (3y^2)(dy/dx). I know that before I take the second derivative I need to solve for y'. But, I don't know how to do this Please Help.

espex · Answer

Solve the equation for dy/dx. $$3y^2\frac{dy}{dx} = -3x^2$$

anonymous · Answer

(dy/dx) = (-x^2/y^2)

espex · Answer

Now you can substitute back and take your second derivative.

anonymous · Answer

so then you would get it down to -3x^4/y^2  - 3y^2x^2/y^2 correct?

espex · Answer

Substitute what you found for dy/dx into the derivative
$$3x^2 + 3y^2\frac{dy}{dx} = 0 \rightarrow 3x^2 + 3y^2\frac{x^2}{y^2} = 0$$

espex · Answer

Correction: That should have been $$-\frac{x^2}{y^2}$$ for dy/dx