Question
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<sup>2</sup> + 2y<sup>2</sup>)<sup>2</sup> - x<sup>2</sup>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))<sup>2</sup> - h(x)*y = C
where f(x) = 4x<sup>2</sup>, g(y)=2y<sup>2</sup> and h(x)=x<sup>2</sup>.
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?
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<sup>2</sup> + 2y<sup>2</sup>)<sup>2</sup> - x<sup>2</sup>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))<sup>2</sup> - h(x)*y = C
where f(x) = 4x<sup>2</sup>, g(y)=2y<sup>2</sup> and h(x)=x<sup>2</sup>.
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?