If sin x - y - logex - y- then dy-dx is equal to

∵ sin(x + y) = log_e (x + y), cos(x + y) ( 1 + dy/dx ) = 1/(x + y) ( 1 + dy/dx ) ⇒ ( 1 + dy/dx ) ( cos ( x + y ) 1/x + y ) = 0 ∵ cos (x + y) 1/(x + y)≠ 0 ...

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Byju's Answer

Standard XII

Mathematics

Implicit Differentiation

If sin x + y ...

Question

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

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Solution

The correct option is D - 1
$∵ s i n (x + y) = l o g_{e} (x + y),$
$c o s (x + y) (1 + \frac{d y}{d x}) = \frac{1}{(x + y)} (1 + \frac{d y}{d x})$
$\Rightarrow (1 + \frac{d y}{d x}) (c o s (x + y) - \frac{1}{x + y}) = 0$
$∵ c o s (x + y) - \frac{1}{(x + y)} \neq 0$
$∴ 1 + \frac{d y}{d x} = 0$
$∴ \frac{d y}{d x} = - 1$

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

The correct option is D - 1∵sin(x+y)=loge(x+y), cos(x+y)(1+dydx)=1(x+y)(1+dydx) ⇒(1+dydx)(cos(x+y)−1x+y)=0 ∵cos(x+y)−1(x+y)≠ 0 ∴1+dydx=0 ∴dydx=−1

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