Find dx dy in the following: ax + by^ 2 = - Vidyadip

Question

Find $\frac{\text{dx}}{{\text{dy}}}$ in the following:
$\text{ax} + \text{by}^{2} = \cos\text{y}$

✓

Answer

The given relationship is $\text{ax} + \text{by}^{2} = \cos\text{y}$
Differentiating this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\text{ax}) + \frac{\text{d}}{\text{dx}}(\text{by}^{2}) =\frac{\text{d}}{\text{dx}}( \cos\text{y})$
$\Rightarrow \text{a + b}\frac{\text{d}}{\text{dx}}(\text{y}^{2}) =\frac{\text{d}}{\text{dx}}( \cos\text{y}) ...(\text{i})$
Using chain rule, we obtain $\frac{\text{d}}{\text{dx}}(\text{y}^{2}) = 2\text{y}\frac{\text{dy}}{\text{dx}}\ \text{and}\ \frac{\text{d}}{\text{dx}}( \cos\text{y}) = -\sin\text{y}\frac{\text{dy}}{\text{dx}} ...{\text{(ii)}}$
Form (i) and (ii), we obtain
$\text{a + b}\times 2\text{y}\frac{\text{dy}}{\text{dx}} = -\sin\text{y}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow(2\text{by} + \sin\text{y})\frac{\text{dy}}{\text{dx}} = -\text{a}$
$\therefore\frac{\text{dy}}{\text{dx}} =\frac{\text{-a}}{2\text{by} + \sin\text{y}}$

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