Question

implicit diff
need help with process (please use dy/dx and dx/dx in notation)
x^2y-3x^2-4=0 at (2,4)

Answer 1

@precal

Answer 2

\[\frac{d}{dx}(x^2y)= y \frac{d}{dx}(x^2) + x^2 \frac{d}{dx}(y) \text{ By product rule! }\]
Do you know how to find
\[\frac{d}{dx} (x^2)=? \text{ and } \frac{d}{dx}(y) =?\]
Well
\[\frac{d}{dx} (y) \text{ is just } \frac{dy}{dx}\]

Answer 3

2x for x^2 (dx/dx) and 1 for y (dy/dx)

Answer 4

\[\text{ so you are saying } \frac{d}{dx}(x^2)=2x \text{ and } \frac{d}{dx}(y)=\frac{dy}{dx}\]
So we have
\[y \cdot 2x+x^2 \cdot \frac{dy}{dx}-\frac{d}{dx}(3x^2)-\frac{d}{dx}(4)=\frac{d}{dx}(0)\]

Answer 5

this is what I got, did I do it right?

Answer 6

\[2xy+x^2 \frac{dy}{dx}-3 \cdot \frac{d}{dx}(x^2)-\frac{d}{dx}(4)=\frac{d}{dx}(0)\]
And you already told me (x^2)'=2x
And I believe you know that the derivative constant is ?

Answer 7

derivative of a constant is 0

Answer 8

and yes your answer is right for finding dy/dx

Answer 9

Ok thanks now I just need to sub x=2 and y=4 into it?

Answer 10

now your question ask you to find dy/dx evalauted at (2,4) ?

Answer 11

Yes

Answer 12

yep ! :)

Answer 13

Thanks you for helping me :)

Answer 14

I like to say
\[\frac{dy}{dx}|_{(2,4)}=-1\]

Answer 15

So they know what point I evaluated y' at