Browse Questions

# Find $\frac {dy}{dx}$ in the following: $xy +y^2 = tan \: x + y$

$\begin{array}{1 1} \frac{sec^2x-y}{x+2y-1} \\ \large \frac{sec^2x+y}{x+2y-1} \\ \frac{cosec^2x-y}{x+2y-1} \\ \frac{cosec^2x+y}{x+2y-1} \end{array}$

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(tanx)}{dx} $$= sec^2x Given xy + y^2 = \tan x + y This is not a standard differentiation of the form y = f(x). We need to differentiate each term separately on LHS and RHS and then calculate \large \frac{dy}{dx} Differentiating both sides: According to the Product Rule for differentiation, given two functions u and v, \large \frac {d(uv)}{dx}$$= u \large \frac{dv}{dx}$$+ v \large \frac{du}{dx}$
$\Rightarrow x\; dy + y\;dx + 2y \; dy = sec^2x \;dx + dy$
$\Rightarrow (x+2y-1)\; dy = (***^2x-y)\; dx$
$\large \frac{dy}{dx} = \frac{sec^2x-y}{x+2y-1}$