If y is a function of x and log(x+y)-2x ...

If `y` is a function of `x` and `log(x+y)-2x y=0,` then the value of `y^(prime)(0)` is
(a)1 (b) `-1` (c) 2 (d) 0

A

1

B

-1

C

2

D

0

Text Solution

Verified by Experts

The correct Answer is:

A

When x=0, we have
`log(0+y)-2xx0xxy=0implieslogy=0impliesy=1`
Now, `log(x+y)-2xy=0`
Differentiating with respect to x, we get
`(1)/(x+y)(1+(dy)/(dx))-2y-2x(dy)/(dx)=0`
Putting x=0 and y=1, we get
`(1+(dy)/(dx))-2-2xx0=0implies(dy)/(dx)=1`

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