Question

find the tangents of the following curves x^2 + 16y^2 = 32

you get\[y'= \frac{-x}{16y}\]please explain

Answer 1

@phi

Answer 2

\[2x+16y^2=0\]\[16y^2=-2x\]\[8y^2=-x\]\[y^2=\frac{-x}{8}\]\[y'=\sqrt{\frac{-x}{8}}dx\]

Answer 3

use implicit differentiation
see http://www.khanacademy.org/math/calculus/differential-calculus/implicit_differentiation/v/implicit-differentiation-1

for
\[ x^2 + 16y^2 =32 \\
\frac{d}{dx} \left(x^2 + 16y^2 =32\right) \\
\frac{d}{dx} x^2 + 16 \frac{d}{dx}y^2 = \frac{d}{dx} 32 \\
2x \frac{dx}{dx} + 16 \cdot 2 y \frac{dy}{dx}= 0
\]
that simplifies to the result.