Posted by **Michael** on Saturday, October 21, 2006 at 7:28pm.

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.

You have

(4x

^{2} + 2y

^{2})

^{2} - x

^{2}y = 4588

It appears you started correctly and you know what you want, dy/dx, but I don't think you did the correct calculations.

It might help to think of this as

(f(x) + g(y))

^{2} - h(x)*y = C

where f(x) = 4x

^{2}, g(y)=2y

^{2} and h(x)=x

^{2}.

Then the differentiation should go

2(f(x) + g(y))*d/dx(f(x) + g(y)) - h(x)*dy/dx -h'(x)y = 0

Then f'(x)=8x, d/dx g(y)=g'(y)*dy/dx and h'(x)=2x

What I suggest, and this what I've done for real long expressions, is to write the symbols out, do the differentiation on the symbols, then substitute them into your equation.

Does this help?

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