Find frac{dy}{dx} for the equation ax + by^2 | vedclass

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(A) The given equation is $ax + by^2 = \cos y$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{

Vedclass · Accepted Answer

$\frac{-a}{2by + \sin y}$

Vedclass · Answer

(A) The given equation is $ax + by^2 = \cos y$. Differentiating both sides with respect to $x$,we get: $\frac{d}{dx}(ax) + \frac{d}{dx}(by^2) = \frac{d}{dx}(\cos y)$ Using the chain rule,we have: $a + b(2y \frac{dy}{dx}) = -\sin y \frac{dy}{dx}$ Rearranging the terms to isolate $\frac{dy}{dx}$: $a = -\sin y \frac{dy}{dx} - 2by \frac{dy}{dx}$ $a = -\frac{dy}{dx}(2by + \sin y)$ Therefore,$\frac{dy}{dx} = \frac{-a}{2by + \sin y}$.

Find $\frac{dy}{dx}$ for the equation $ax + by^2 = \cos y$.

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