Question

Evaluate the derivative dy/dx at the (0,0) for the equation 2x -5x^3y^2+4y=0

Answer 1

thats implicit derivative.
do you know product and chain rules ?

Answer 2

So I can just use the product rule?

Answer 3

you will need both product and chain rule for the term 5x^3y^2

Answer 4

let me give you example
derivative of xy w.r.t x will be
x (dy/dx) + y (dx/dx) = x dy/dx + y

Answer 5

so is it 0= 2-5x^3* 2y -

Answer 6

Sorry not that
computer sent it when I wasnt done

Answer 7

0= 2-5x^3 * 2y -15x^2 * y^2 + 4

Answer 8

note :
derivative of y^2 with respect to 'x' will require chain rule~

d/dx (y^2) = 2y dy/dx
see if u get this ?

Answer 9

similarly for 4y
d/dx (4y) = 4 dy/dx

(since we have 'y' but we are differentiating with respect to 'x', thats why we need chain rule)

Answer 10

So wait it then becomes 2-5x^3 * 2 dy/dx - 15x^2 + 4 dy/dx

Answer 11

If its not that, then Im completely lost and have no clue

Answer 12

just one error, maybe typing,
" 2-5x^3 * 2y dy/dx - 15x^2 + 4 dy/dx =0 "

Answer 13

there should have been 2y dy/dx instead of 2 dy/dx
because
d/dx (y^2) = 2y dy/dx

Answer 14

got it ?

Answer 15

Oh okay yeah! I got it. So then do you leave it like that? or is the next step to make it all equal to dy/dt?

Answer 16

we need to evaluate the derivative dy/dx AT (x,y) = (0,0)
so, you just need to plug in x =0 and y = 0 in
2-5x^3 * 2y dy/dx - 15x^2 + 4 dy/dx = 0
and isolate dy/dx

Answer 17

Sorry at last,there is a word 'near'.I didn't notice it before posting.

Answer 18

ohh okay thank you

Answer 19

No problem...