Question

Find Dy/dx when x^3e^y - y^2 e^x = 2 .

Answer 1

its just implicits ....

Answer 2

product rule, power rule, "e" rule ... you know the derivative rules right?

Answer 3

yes, but I'm confused:/

Answer 4

i know, they start you out with x and y as if there is something special about the names ... there really isnt

Answer 5

I solved it , but the answer is wrong:x

Answer 6

\[D[x^3e^y - y^2 e^x = 2]\]
\[D[x^3e^y] - D[y^2 e^x] = D[2]\]
\[D[x^3]e^y+x^3D[e^y] - D[y^2] e^x- y^2 D[e^x] = 0\]
\[3x^2x'e^y+x^3y'e^y - 2yy' e^x- y^2 x'e^x = 0\]
since we want to define this for dy/dx, let dx/dx=1
\[3x^2e^y+x^3y'e^y - 2yy' e^x- y^2 e^x = 0\]
solve for y'

Answer 7

no matter what the name of the variable, the chain rule will always pop out a '

thats all implicits are is application of the chain rule. the rest is algebra :)

Answer 8

I didn't study dx/dx..

Answer 9

what should I do if I got stuck?

Answer 10

tomorrow's final man:')

Answer 11

recall that dy/dx is the relationship between how fast y moves as x moves .... the slope of the line at a given point

dx/dx is the relationship between how fast x moves as x moves ..... its going to always be a 1:1 ratio

Answer 12

what part is giving you the trouble? derivatives, or algebra?

Answer 13

oh I got it. I have no trouble:p but sometimes I got stuck and dunno how to differentiate:)

Answer 14

consider y=x :)

D[y = x]

D[y] = D[x]

y' = x' , if our goal is wrt.x; then x' = dx/dx, and y' = dy/dx

y' = 1

Answer 15

great:D thaaanks:D

Answer 16

good luck :)