Question

if y is a function of x. and log(x+y)-2xy=0 what is the value of f'(0)?

Answer 1

\[\begin{array}{rcl}
\ln(x+y)-2xy&=&0\\
\ln(x+y)&=&2xy\\
\frac{1+\frac{dy}{dx}}{x+y}&=&2y+2\frac{dy}{dx}\\
\end{array}\]

Answer 2

then i have no idea xP

Answer 3

i did upto this.

Answer 4

theres a mistake in your derivative

Answer 5

log(x+y)-2xy=0
log (x+y) = 2xy
(1 + y' ) / (x+y) = 2*1*y + 2*x*y'

Answer 6

what was the mistake?

Answer 7

your right side, should be 2y + 2xy'

Answer 8

oh i see

Answer 9

now substitute y'(0) , , given that x = 0, and y = (you can solve for this)

Answer 10

mustafa made the same mistake -sigh-

Answer 11

at x=0 y=1

Answer 12

log(x+y)-2xy=0
log (x+y) = 2xy
(1 + y' ) / (x+y) = 2*1*y + 2*x*y'

Answer 13

okay

Answer 14

when x = 0, then
log( 0 + y ) - 2(0)*y = 0
log y - 0 = 0
log y = 0
y = 1.
now plug in , x = 0, y = 0, and solve for y'(0)

Answer 15

(1 + y'(0) ) / (0+1) = 2*1*1 + 2*0*y'(0)

Answer 16

( 1 + y'(0) ) /1 = 2
y'(0) = 1

Answer 17

thanks