Question

Find dx/dy for the equation xy = -12.

Answer 1

you could solve for x
and like x=-12/y
and then take derivative of both sides with respect to y

Answer 2

you could rewrite -12/y as -12y^(-1)
and use power rule

Answer 3

I don't really get how to find derivative, my teacher left and our substitute is terrible. I sort of understand power rule

Answer 4

power rule is
\[(x^n)'=n x^{n-1}\]

here we just have a different variable y instead of x
and our n is -1 in this case since we have y^(-1)

-12 is just a constant multiple...ignore it and bring it down after application of power rule

Answer 5

\[\frac{d}{dy} y^{\color{red}{n}}=\color{red}n y^{\color{red}n-1} \\ \frac{d}{dy}y^{\color{red}{-1}}=? \]

Answer 6

Sorry my computer's being slow.

would it be d/dy y^-1 = -1y^-1-1?

Answer 7

A key fact here: You have a composite function in xy and thus need to use the Power Rule AS WELL AS the Chain Rule to differentiate xy.

xy=-12 could also be solved for y, as myininaya has pointed out.

In that case, y=-12/x. I agree with that.

Let's try finding the derivative dy/dx if y=-12/x or y=-12x^(-1).

dy/dx-12 * (d/dx)(x^(-1) ), Can you finish this? See myininaya's latest post, above.



If you wish, you can differentiate xy=-12 implicitly (without first having solved for y).

First apply the Product Rule:\[\frac{ d }{ dx }xy=\frac{ d }{ dx }(-12) = 0\]

Answer 8

\[\frac{ d }{ dx }xy=\frac{ d }{ dx }(-12) = 0\]

Answer 9

Let's go back to the "solve the equation for y first" appraoch.

"would it be d/dy y^-1 = -1y^-1-1? " You're on the right track, but a little housekeeping is necessary here.

You want to find \[\frac{ d }{ dx }y ^{-1}\]

Answer 10

whereas you were differentiating with respect to y.

Otherwise you were correct, but did not complete the differentiation

Answer 11

\[\frac{ d }{ dx }y ^{-1}=-y ^{-2}\frac{ dy }{ dx }\]

Answer 12

ohhhh. that makes more sense

Answer 13

we are asked to find dx/dy
this is why i differentiated with respect to y and not x

Answer 14

That, actually, is the answer to this problem.

Note you could also use implicit differentiation, as mentioned above, and would most likely obtain the same answer.

@myininaya: You are indeed correct.

I'm going to defer to you at this point. Mind guiding Love the rest of the way through this problem? Thx.

Answer 15

\[\frac{d}{dy}y^{-1}=-1y^{-1-1} \text{ is correct though } \\ \text{ the only cleanup needed is } \\ \frac{d}{dy}y^{-1}=-1y^{-2}=\frac{-1}{y^2}\]

now going back to
\[x=\frac{-12}{y}\]
you can write as
\[x=-12 \cdot \frac{1}{y} \text{ or } x=-12 y^{-1} \]
write your d/dy on both sides
\[\frac{d}{dy}(x)=\frac{d}{dy}(-12 y^{-1}) \\ \frac{dx}{dy}=-12 \frac{d}{dy}(y^{-1}) \text{ since } -12 \text{ is a constant multiply }\]
you just need to replace d/dy (y^(-1) with what you found above
and you are done

Answer 16

replace it with -y^-2?

Answer 17

yep
\[\frac{dx}{dy}=-12(-y^{-2})\]
you could do a little more... sort of
like the neg*neg part
and also write your dx/dy expression without negative exponents

Answer 18

Can't we just divide both sides by x, and differentiate explicitly ??

Answer 19

xy=-12 ---> y=-12/x ---> dy/dx = (-12x\(^{-1}\))\({\tiny~}'\)

Answer 20

I will give you an example.

Answer 21

Find \(\color{black}{dy/dx}\) for \(\color{black}{x^2y=16}\).

------------------------------------------------------------
First, we will differentiate both sides wrt \(\color{black}{x}\). \(\tiny \\[0.7em]\)
\(\color{black}{\displaystyle x^2\frac{dy}{dx}+2xy=0 \quad \Longrightarrow \quad x^2\frac{dy}{dx}=-2xy \quad \Longrightarrow \quad \frac{dy}{dx}=-\frac{2y}{x}}\)

We know that \(\color{black}{y=16/x^2}\) (basic one-step algebraic manipulation), thus,
\(\color{black}{\displaystyle \frac{dy}{dx}=-\frac{2(16/x^2)}{x}=-\frac{32}{x^3}}\).

Alternatively, we can solve for \(\color{black}{y}\) and use the power rule.
\(\color{black}{\displaystyle x^2y=16\quad \Longrightarrow \quad y=16x^{-2}\quad \Longrightarrow \quad\frac{dy}{dx}=-32x^{-3}}\)

Answer 22

You need to know the first method, though, in case you can't solve for \(y\),
(e.g. \(\color{black}{e^y+yx=14x-y^4}\) )

Answer 23

but, if you can easily solve for \(y\), then you might as well do it by solving for y, and then using the power rule. (Then, you can confirm that you know the approach where you differentiate without solving for y by getting the same answer.)

Answer 24

did you get the answer you needed