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If y is a function of x and log (x+y)=xy then y` (0) = |
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Answer» When x=0, we have log(0+y)−2×0×y=0 ⇒logy=0 ⇒y=e 0 =1 Given equation is log(x+y)−2xy=0 On differentiating w,r,t x we get x+y 1
(1+ dx dy
)−2y−2x dx dy
=0 On PUTTING x=0,y=1 we get (1+ dx dy
)−2−0=0 ⇒(1+ dx dy
)−2=0 ⇒ dx dy
=1 ie,y ′ (x)=1 ⇒y ′ (0)=1 |
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