Question

Differentiate implicitly:

(x^2+y^2)=(2x^2)-2y^2-x)^2

I have it mostly worked out, just somehow my answer is negative when it should be positive. I'll post more

Answer 1

\[x ^{2} + y ^{2}=(2x ^{2}+ 2y ^{2} - x) ^{2}\]
Is the equation

Oh yeah, I'm supposed to figure out the equation of the tangent line at the point (0, 1/2)

Answer 2

\[8x ^{3} - 6x ^{2} + 8xy ^{2} - 2y ^{2} \over y - 8x ^{2} - 8y ^{3} + 4xy\] Is the derivative I got

Answer 3

And then plugging in the point (0, 1/2) I get a slope of -1

Answer 4

1. The parenthesis are not balanced in the problem equation.
The following appears to be the problem equation:\[x^2+y^2=\left(2 x^2-x-2 y^2\right)^2 \]

Answer 5

The equation I have in the first reply is the equation in the book.

Answer 6

I guess my problem is, I got a negative 1 slope even though it's supposed to be +1 and I don't see where I got it wrong

Answer 7

\[x^{2} + y^{2} = (2x^{2}+2y^{2}-x)^{2}\] \[2x + 2yy' = 2(2x^{2} + 2y^{2}-x)(4x+4yy'-1)\] I wont bother calclutaing y' in terms of x,y just plug it in at this step itself,
\[2(1/2)y'= 2(2(1/2)^{2})(4(1/2)y'-1) \] gives
\[y'= 2y' - 1\] or y' = 1

Answer 8

Wow, your method is much simpler. Thanks!

Answer 9

For one who cannot manually find the derivative of x:
A problem solution using Mathematica 8 is attached.

Answer 10

The above was re-posted to correct a reference error in line In[2]. The calculations were not effected.