If sin x-y-log x-y- then d y d x-a 2b -2cd - 1-

(d) minus; 1 We #160;have, #160;sinx+y=logx+y #8658;cosx+y1+dydx=1x+y1+dydx #8658;cosx+y+cosx+ydydx=1x+y+1x+ydydx #8658;cosx+ydydx 1x+ydydx=1x+y c ...

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Byju's Answer

Standard XII

Mathematics

Monotonicity in an Interval

If sin x+y=lo...

Question

If $\sin (x + y) = \log (x + y), then \frac{d y}{d x} =$
(a) 2
(b) − 2
(c) 1
(d) − 1]

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Solution

(d) − 1

$We have, \sin (x + y) = \log (x + y) \Rightarrow \cos (x + y) (1 + \frac{d y}{d x}) = \frac{1}{(x + y)} (1 + \frac{d y}{d x}) \Rightarrow \cos (x + y) + \cos (x + y) \frac{d y}{d x} = \frac{1}{(x + y)} + \frac{1}{(x + y)} \frac{d y}{d x} \Rightarrow \cos (x + y) \frac{d y}{d x} - \frac{1}{(x + y)} \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \Rightarrow \{\cos (x + y) - \frac{1}{(x + y)}\} \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \Rightarrow - \{\frac{1}{(x + y)} - \cos (x + y)\} \frac{d y}{d x} = \frac{1}{(x + y)} - \cos (x + y) \Rightarrow \frac{d y}{d x} = - 1$

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If sin x+y=log x+y, then dydx= (a) 2 (b) − 2 (c) 1 (d) − 1]

(d) − 1 We have, sinx+y=logx+y⇒cosx+y1+dydx=1x+y1+dydx⇒cosx+y+cosx+ydydx=1x+y+1x+ydydx⇒cosx+ydydx-1x+ydydx=1x+y-cosx+y⇒cosx+y-1x+ydydx=1x+y-cosx+y⇒-1x+y-cosx+ydydx=1x+y-cosx+y⇒dydx=-1

If $\sin (x + y) = \log (x + y), then \frac{d y}{d x} =$
(a) 2
(b) − 2
(c) 1
(d) − 1]