Implicit Differentiation: Let x^3 + y^3 = 28. Find y"(x

[chucknorrisfish]

1) Let x^3 + y^3 = 28. Find y"(x) at the point (3, 1). y"(3)=

The correct answer is: -2*3*28 = -168

3x^2 + 3y^2(dx/dy) = 0

6x + 6y(dx/dy) = 0

dx/dy = -6(3)/6(1) = -3

That's as far as i get... i don't even know if that's started right...

One more thing confuses me:

2) Regard y as independent variable and x as dependant variable and find the slope of the tangent line to the curve (4x^2 + 2y2)^2 - x^2y = 4588 at point (3,4).

Correct answer is -0.668827160493827

Here's what I did:

2(8x(dy/dx) + 4y) -2x(dx/dy)y + x^2 = 0
16x(dx/dy) + 8y - 2x(dx/dy)y = -x^2
16x(dx/dy) - 2x(dx/dy)y = -x^2 - 8y
(dx/dy)(16x -2xy) = -x^2 - 8y
(dx/dy) = (-x^2 - 8y)/(16x -2xy)

But this doesnt work.

 

[tkhunny]

Re: Implicit Differentiation

3x^2 + 3y^2(dx/dy) = 0

There's the first problem. Why is it dx/dy rather than dy/dx?

dy/dx = =3x^2/3y^2 = -x^2/y^2

Use the Quotient Rule to find the next derivative.

OR

3x^2 + 3y^2(dy/dx) = 0

Use the product and chain rules to find the next derivative.