Question

Using implicit Differentiation to find the second derivative of dy/dx=-x/y

Answer 1

Use the quotient rule to find the derivative of -x/y.

Answer 2

i did, but now i;m stuck on a part

Answer 3

Which part?

Answer 4

i have this so far\[\left(\begin{matrix}-y+x(-x/y) \\ y^2\end{matrix}\right)\]

Answer 5

I think you are missing a negative
Like pretend if we just (x/y)'=
\[=\frac{1y-xy'}{y^2}=\frac{y-x \frac{-x}{y}}{y^2}\]
(-x/y)'=-(x/y)'=
\[-\frac{y+\frac{x^2}{y}}{y^2}\]
Multiply top and bottom by y.

Answer 6

my answer book has it like that, and then does other steps

Answer 7

Like multiplying it by y/y

Answer 8

You do this to clear the compound fractions.

Answer 9

i think so. idk what those r . i never did this before and my prof didnt go over it

Answer 10

r?

Answer 11

Compound fractions you mean?

Answer 12

That is when there is a fraction inside of a fraction.

Answer 13

no i mean taking the second derivative for implicit Differentiation

Answer 14

To find the second derivative you just find the derivative of the expression that is the first derivative.

Answer 15

yes i know that part, but im kinda stuck on what to do at one step. idk if i should multiply the fraction of -x/y with -y+x or just x

Answer 16

Are you talking to clear the compound fraction?

Answer 17

yea

Answer 18

You multiply by y/y

Answer 19

there is a y in the bottom of that fraction inside the fraction
so multiplying it by y will help us deal with the y in the bottom
if we multiply y on top then we have to multiply y on bottom.

Answer 20

ok, im confused now. bc i have -x/y on the top multiplied by (-y+x). if I multiply by y/y then the +x will be xy. and i need to get -y^2+x^2

Answer 21

we have
\[-\frac{y+\frac{x^2}{y}}{y^2} \cdot \frac{y}{y}= -\frac{y \cdot (y+\frac{x^2}{y})}{y \cdot y^2 } \]

Answer 22

Distribute,

Answer 23

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

Answer 24

oh, i was doing it the wrong way then