Question

Derivative x^-2/y

Answer 1

So implicit differentiation, eh?

Answer 2

Yes

Answer 3

Ah, okay. Well, we still want to treat this like a normal derivative, but the difference is we want to pay attention and mark when we take a derivative of y. So we start with quotient rule.

Answer 4

\[\frac{ (y)(-2x ^{-3})-(x ^{-2})(1\frac{ dy }{ dx } }{ y }\]

Answer 5

y^2

Answer 6

Yeah, you saw it, haha. Right, so now we want to separate the derivative of y from everything else. The derivative of y is multiplying into the x^-2 term, so that is the term we have to isolate by itself on one side of the equation. So see if you can rearrange everything as to where x^-2dy/dx is by itself on one side.

Answer 7

\[y ^{2}=y(-2x ^{-3})-x ^{-2}(\frac{ dy }{ dx }\]

Answer 8

Well we want the dy/dx term by itself. So we want x^(-2)dy/dx by itself. This is how I would go about doing it:
\[\frac{ (y)(-2x ^{-3})-(x ^{-2})(\frac{ dy }{ dx }) }{ y ^{2} }=\frac{ -2x ^{-3} }{ y }=\frac{ (x ^{2})(\frac{ dy }{ dx }) }{ y ^{2} }\]
So now all you need to do is get dy/dx by itself completely. Once you do that youre done :3

Answer 9

Oops, forgot to make the x^2 a x^(-2), lol.

Answer 10

\[y ^{2}(\frac{ -2x ^{-3} }{ y })=x ^{2}\]

Answer 11

Yeah, let me make that correction for the ^-2

\[y ^{2}(\frac{ -2x ^{-3} }{ y })=x ^{-2}\frac{ dy }{ dx } \]
Now we;re not done. We want dy/dx completely by itself.

Answer 12

\[x ^{2}y ^{2}(\frac{ -2x ^{-3} }{ y })=\frac{ dy }{ dx }\]

Answer 13

Right, now just simplify and ya got it :3

Answer 14

\[\frac{ -2xy ^{-3}-2x ^{4} }{ y ^{5} }\]

Answer 15

Not quite sure how you got that. We would have:
\[\frac{ -2x ^{2}x ^{-3}y ^{2} }{ y }\]