Question

If $Sin (x + y)+Cos(x + y) = Log (x + y),$ then $\frac {d^2y}{dx^2}$=

• A. 1
• B. -1
• C. 0
• D. $\frac {-y}{x}$

Answer 1

Answer: (C)

0

Answer 2

We have
$\sin (x+y)+\cos (x+y)=\log (x+y)$
On differentiating both sides, we get
$\cos (x+y)\left(1+\frac{d y}{d x}\right)-\sin (x+y)\left(1+\frac{d y}{d x}\right)$
$=\frac{1}{x+y}\left(1+\frac{d y}{d x}\right)$
\Rightarrow \frac{d y}{d x}+1=0 \\\
$\Rightarrow \frac{d y}{d x}=-1$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=0$