Question

If $\sin(x + y) = \log(x + y)$, then $\frac{dy}{dx}$=

• A. 2
• B. – 2
• C. 1
• D. – 1

Answer 1

Answer: (D)

– 1

Answer 2

$\sin(x + y) = \log(x + y)$

Differentiating with respect to x,

$\cos ( x + y ) \left[ 1 + \frac { d y } { d x } \right] = \frac { 1 } { x + y } \left[ 1 + \frac { d y } { d x } \right]$ $\left\lbrack \cos(x + y) - \frac{1}{x + y} \right\rbrack\left\lbrack 1 + \frac{dy}{dx} \right\rbrack = 0$

$\because$ $\cos(x + y) \neq \frac{1}{x + y}$ for any x and y. So, $1 + \frac{dy}{dx} = 0$,

$\frac{dy}{dx} = - 1$.

Trick: It is an implicit function, so

$\frac{dy}{dx} = - \frac{\partial f/\partial x}{\partial f/\partial y} = - \frac{\cos(x + y) - \frac{1}{x + y}}{\cos(x + y) - \frac{1}{x + y}} = - 1$.