Question

Find the slope at x^2y + y^2 = 5 at points ( 2,1)

I keep getting 1/3. I don't think that is the right answer though..

Answer 1

u mean the slope of the tangent line?

Answer 2

Idk, it just says find the slope of that equation. I think you have to use implicit differentiation to find the answer

Answer 3

yes u have to differentiate implicitly:
\[\large 2xy+x^2y'+2yy'=0 \]
\[\large -2xy=y'(x^2+2y) \]
\[\large y'=\frac{-2xy}{x^2+2y} \]
did u get the same?

Answer 4

umm no.. I got -2/ 2x + 2

Answer 5

how come?

Answer 6

\[( 2x dy/dx + 2y) + 2ydy/dx = 0 \]

Answer 7

then I just factor out dy/dx

Answer 8

NO. the first term is a produc
\[\large (uv)'=u'v+uv' \]

Answer 9

Yea I know. So I did f = 2x
g = y
f' = 2
g' = dy/dx
Then I multiplied ( f.g') + (g.f')

Answer 10

but you wrote \(x^2y\) originally

Answer 11

yea, but when you take the derivative of that won't it gives you 2xydy/dx? Then you would take that and put it into the product rule? I don't even know

Answer 12

yes, but when u differentiate \(x^2\) u DO NOT differentiante \(y\) as well

Answer 13

So can you show me how to take the derivative of that equation ? and put it in product rule?

Answer 14

ok

Answer 15

Thanks!

Answer 16

\[\large x^2y+y^2=5 \]
differentiating implicitly (using the chaing rule for \(y^2\))
\[\large (x^2y)'+(y^2)'=(5)' \]
\[\large (x^2)'(y)+(x^2)(y)'+2yy'=0 \]
\[\large 2xy+x^2y'+2yy'=0 \]
solving for \(y'\):
\[\large y'=\frac{-2xy}{x^2+2y} \]

Answer 17

Thank you so much. I got -2/3 as the answer.

Answer 18

yes, -2/3 is right.

and u r welcome