MCQ
If $sin(x + y)$ + $cos(2x + 2y)$ = $ln(3x + 3y)$, then $\frac{{dy}}{{dx}}$ is
- A$-2$
- ✓$-1$
- C$2$
- D$1$
$=\frac{1}{3(x+y)} \cdot\left(3+\frac{3 d y}{d x}\right)$
$\Rightarrow\left(1+\frac{d y}{d x}\right)\left(\cos (x+y)-2 \sin (2 x+2 y)-\frac{1}{x+y}\right)=0$
$\Rightarrow \frac{d y}{d x}=-1$
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$[A]$ $f(x)$ is increasing in $(0, \infty)$
$[B]$ $f(x)$ is decreasing in $(0, \infty)$
$[C]$ $f(x)>e^{2 x}$ in $(0, \infty)$
$[D]$ $f^{\prime}(x) < e^{2 x}$ in $(0, \infty)$