Saturday
April 19, 2014

Homework Help: Calc, Implicit Differentiation

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
(4x2 + 2y2)2 - x2y = 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) = 4x2, g(y)=2y2 and h(x)=x2.
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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