Question

can somebody explain me implicit differentiation
with this problem

xy^3 + xy = 4

Answer 1

@Zarkon

Answer 2

\[\begin{align*}\frac{d}{dx}\left[xy^3+xy\right]&=\frac{d}{dx}4\\
\frac{d}{dx}[xy^3]+\frac{d}{dx}[xy]&=0\\
\frac{d}{dx}[x]y^3+x\frac{d}{dx}[y^3]+\frac{d}{dx}[x]y+x\frac{d}{dx}[y]&=0\\
y^3+x\left(3y^2\frac{dy}{dx}\right)+y+x\frac{dy}{dx}&=0
\end{align*}\]

Answer 3

Solve for \(\dfrac{dy}{dx}\)

Answer 4

@dan815

Answer 5

@SithsAndGiggles could you explain it to me in words, please and thank you

Answer 6

i got the answer but it is so difficult to explain implicit diff. Hopefully @zzr0ck3r can do a fair job of it!

Answer 7

the answer is DAK DAK DAK DAK DAK DAK DAK

Answer 8

line 1: take derivative of both sides
line 2:use the fact that differentiation is a linear operator so you can "split" things apart
line3:use the product rule for differentiating
line 4:take all the derivatives

note: since y depends on x, when we take the derivative of 2y you get 2 y' and the derivative of y^2 = 2yy' because of the chain rule

Answer 9

chain rule, since y=f(x)
so if you have g(y)
then dg/dx= g'(y)*dy/dx

Answer 10

and holy god is @SithsAndGiggles do a good job of making it look pretty

Answer 11

@zzr0ck3r thanks!

Answer 12

agreed lol