Using implicit differentiation, find the derivative

y = -ln(4x^2 + 8y^2)

What I got so far:
y' = -(8x+16yy')/(4x^2 + 8y^2)

How would I solve for y' algebraically?

Thanks!

Question

Using implicit differentiation, find the derivative

y = -ln(4x^2 + 8y^2)

What I got so far:
y' = -(8x+16yy')/(4x^2 + 8y^2)

How would I solve for y' algebraically?

Thanks!

zale101 · Answer

$y = -ln(4x^2 + 8y^2)$

$y'=\large -\frac{1}{4x^2+8y^2}*8x+16y*y'$

$y'=\LARGE -\frac{8x+16y*y'}{4x^2+8y^2}$

Factor out the y' by multiplying both sides by 4x^2+8y^2, then expand it.

zale101 · Answer

$\Large y'=\frac{16y*y'-8x}{4x^2+8y^2}$

$ (4x^2+8y^2)y'=\Large{\frac{16y*y'-8x}{4x^2+8y^2}}*(4x^2+8y^2)$

$ (4x^2+8y^2)y'=16y*y'-8x$

distribute the y' in then do some further simplifying and canceling

zale101 · Answer

Clear? @brando86

anonymous · Answer

Ah, yes I understand it now. Thanks!

zale101 · Answer

No problem! Tag me when you get stuck at something.