Question

Just need to make sure.
evaluate the derivative dy/dx at the point (0,0), for the equation 2x-5x^3y^2+4y=0

Answer 1

@RBauer4

Answer 2

i think it is undefined

Answer 3

my options are 3, -1/2, undefinded, 6, and none of these

Answer 4

hmm what did you get for the \(\cfrac{dy}{dx}\) of \(2x-5x^3y^2+4y=0\) ?

Answer 5

I got 2-15x^2y^2

Answer 6

2x - 5x^3y^2 + 4y
^ ^ ^
power rule product rule power rule

Answer 7

i dont understand

Answer 8

@jdoe0001

Answer 9

\(\bf 2x-5x^3y^2+4y=0\qquad \cfrac{dy}{dx}\implies 2-(15x^2\cdot y^2+5x^3\cdot 2y)+4\\ \quad \\
\cfrac{dy}{dx}\implies -15x^2y^2-10x^3y+6\)

Answer 10

oh ok i did it wrong :/

Answer 11

so now I insert 0 in x and y?

Answer 12

well... that's the derivative equation, the equation for the slope, now to get the slope itself at (0, 0), just set x =0

that'd yield a "y" of 6

Answer 13

so the answer is 6? ... wow I did that very wrong

Answer 14

yes

Answer 15

thank you so much. :)

Answer 16

yw

Answer 17

could you help me with one more?