MCQ
If $sin(x + y)$ + $cos(2x + 2y)$ = $ln(3x + 3y)$, then $\frac{{dy}}{{dx}}$ is
  • A
    $-2$
  • $-1$
  • C
    $2$
  • D
    $1$

Answer

Correct option: B.
$-1$
b
$\cos (x+y)\left(1+\frac{d y}{d x}\right)-\sin (2 x+2 y)\left(2+\frac{2 d y}{d x}\right)$

$=\frac{1}{3(x+y)} \cdot\left(3+\frac{3 d y}{d x}\right)$

$\Rightarrow\left(1+\frac{d y}{d x}\right)\left(\cos (x+y)-2 \sin (2 x+2 y)-\frac{1}{x+y}\right)=0$

$\Rightarrow \frac{d y}{d x}=-1$

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