Browse Questions

# Find $\frac {dy}{dx}$ in the following: $2x +3y = sin \: y$

$\begin{array}{1 1} \large \frac {2}{cosy-3} \\ \large \frac {2}{cosy+3} \\ \large \frac {-2}{siny+3} \\ \large \frac {2}{siny-3} \end{array}$

Toolbox:
• For equations that are not of the form $y = f(x)$, we need to differentiate each term separately on LHS and RHS and then calculate $\large \frac{dy}{dx}$
• $\; \large \frac{d(cosx)}{dx} $$=-sinx Given 2x +3y = sin \: y: This is not a standard differentiation of the form y = f(x). We need to differentiate each term separately on LHS and RHS and then calculate \large \frac{dy}{dx} \; \large \frac{d(cosx)}{dx}$$=-sinx$
Differentiating both sides, $\rightarrow 2\; dx + 3\; dy = cosy\; dy$
$\Rightarrow dy \; (cosy - 3) = 2\; dx$
$\Rightarrow \large \frac{dy}{dx} = \frac {2}{cosy-3}$