If sin (x+y)=loge(x+y),thendydx is equal to

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Question

I f s i n (x + y) = l o g_{e} (x + y), t h e n \frac{d y}{d x} i s e q u a l t o

Options:

A . 2

B . - 2

C . 1

D . - 1

Answer: Option D
:
D

∵ s i n (x + y) = l o g_{e} (x + y),

c o s (x + y) (1 + \frac{d y}{d x}) = \frac{1}{(x + y)} (1 + \frac{d y}{d x})

\Rightarrow (1 + \frac{d y}{d x}) (c o s (x + y) - \frac{1}{x + y}) = 0

∵ c o s (x + y) - \frac{1}{(x + y)} \neq 0

∴ 1 + \frac{d y}{d x} = 0

∴ \frac{d y}{d x} = - 1

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