Find d2y/dx2 by implicit differentiation, given x^2 + 3y^2 = 5. Help please!
step number 1 take the derivative wrt x you get \[2x+6yy'=0\] is that part ok?
yes I got that
then solve for \[y'\] via \[6yy'=-2x\] \[y'=-\frac{x}{3y}\]
Mhm I got that far
so far so good?
Yes :)
ok so now we do it again, this time with the paininthebutt quotient rule
OOOOH
want to factor out the \[-\frac{1}{3}\] first or leave it in?
we can leave it in I guess. Which ever one is easier...
lets make life easy and factor it out so we have \[-\frac{1}{3}\times \frac{x}{y}\] and we can mutiply by \[-\frac{1}{3}\] at the end
ok :)
so now for \[y''=-\frac{1}{3}\times \frac{y-xy'}{y^2}\]
we are not done yet, but is that ok?
yes :)
ok and last step is an algebra one. because we actually know what \[y'\] is
\[y'=-\frac{x}{3y}\]so that is what we replace it by in our second derivative to get \[y''=-\frac{1}{3}\times \frac{y-x\times -\frac{x}{3y}}{y^2}\]
still not done. more algebra
ok >.<
need to clear the compound fraction, so maybe multply top and bottom by \[3y\] to get \[y''=-\frac{3y^2+x^2}{9y^3}\]
and we are STILL NOT DONE
because in a fit of trickery they have arranged it so that you numerator is in fact what you started with (be on the lookout for this) recall that we started with \[x^2+3y^2=5\] so the numerator is just... 5! and you get \[y''=-\frac{5}{9y^3}\] (check my algebra)
You... are a genius. I was about to cry. Thank you sooo much!
actually myininay will check my algebra, she loves finding my mistakes
ooops i got a medal. i guess it is right
You were completely correct! Thank you again!
well there is one thing i don't like that you said...
yw
i just don't like \[\times \] this symbol
but goof job
i mean good job lol
oh i thought you were going to tell me i was off by a minus sign or something. i put that in to make sure it is clear that we were multiplying. i like "goof job" better actually
me too
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