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

Question

kc_kennylau · Answer

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

kc_kennylau · Answer

then i have no idea xP

random231 · Answer

i did upto this.

perl · Answer

theres a mistake in your derivative

perl · Answer

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

kc_kennylau · Answer

what was the mistake?

perl · Answer

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

kc_kennylau · Answer

oh i see

perl · Answer

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

perl · Answer

mustafa made the same mistake -sigh-

anonymous · Answer

at x=0   y=1

perl · Answer

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

random231 · Answer

okay

perl · Answer

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)

perl · Answer

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

perl · Answer

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

random231 · Answer

thanks