Question
Find $\frac{{dy}}{{dx}}$ if $xy + {y^2} = \tan x + y$
$\Rightarrow \frac{d}{{dx}}\left( {xy} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) = \frac{d}{{dx}}\tan x + \frac{d}{{dx}}y$
$\Rightarrow x\frac{dy}{{dx}} + y.1 + 2y\frac{dy}{{dx}} = {\sec ^2}x + \frac{dy}{{dx}}$ [By Product Rule]
$\Rightarrow x\frac{dy}{{dx}} + 2y\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = {\sec ^2}x - y$
$$$\Rightarrow \left( {x + 2y - 1} \right)\frac{{dy}}{{dx}} = {\sec ^2}x - y$
$$$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{{\sec }^2}x - y}}{{x + 2y - 1}}$
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| $X_i$ | $0$ | $1$ | $2$ |
| $P_i$ | $3c^3$ | $4c - 10c^2$ | $5c - 1$ |