Find the tangent line to the equation:

x^4 = 4(x^2 y^2) at [1, sqrt(3/2) ]

I got the derivative as (8y-4x^3)/8y

Trying to simplify that seems to give me different slope, and the answer in the book is wrong. The book says the slop should be sqrt(3)

Question

Find the tangent line to the equation:

x^4 = 4(x^2 y^2) at [1, sqrt(3/2) ]

I got the derivative as (8y-4x^3)/8y

Trying to simplify that seems to give me different slope, and the answer in the book is wrong.  The book says the slop should be sqrt(3)

berrymox · Answer

$$y' = \frac{ 8x-4x ^{3} }{ 8y }$$

$$f(x): x ^{4} = 4(x ^{2} y^{2})$$

zepdrix · Answer

Hmm this problem is strange...
Why not just divide both sides by x^2,
$\large\rm x^2=4y^2,\qquad\qquad x\ne0$
And differentiate from there.
Thinking...

zepdrix · Answer

Anyway, doing it the hard way :)
$\large\rm 4x^3=8xy^2+8x^2yy'$
Is this what you ended up with after differentiating implicitly?

nincompoop · Answer

can you show your work?

robtobey2 · Answer

Refer to the Mathematica attachment.