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  • Find dy/dx when x and y are connected by the relation given sin(xy) + x/y = x^2-y

Find \frac{dy}{dx} when x and y are connected by the relation given

\sin (x y)+\frac{x}{y}=x^{2}-y

Answers (1)

We have,

\sin (x y)+\frac{x}{y}=x^{2}-y

Use chain rule and quotient rule to get:

Chain Rule

f(x) = g(h(x))

f’(x) = g’(h(x))h’(x)

By Quotient Rule

\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left[\frac{\mathrm{f}(\mathrm{x})}{\mathrm{g}(\mathrm{x})}\right]=\frac{\mathrm{g}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})-\mathrm{f}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})}{[\mathrm{g}(\mathrm{x})]^{2}}

On differentiating both the sides with respect to x, we get

\begin{aligned} &\cos x y \times \frac{d}{d x}(x y)+\frac{y \frac{d}{d x}(x)-x \frac{d}{d x}(y)}{y^{2}}=2 x-\frac{d y}{d x}\\ &\text { By product rule: }\\ &\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})]=\mathrm{f}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})\\ &\left[\because \frac{d}{d x} \sin x=\cos x\right] \end{aligned}

\begin{aligned} &\Rightarrow \cos (x y)\left[x \frac{d y}{d x}+y \times(1)\right]+\frac{y \times 1-x \frac{d y}{d x}}{y^{2}}=2 x-\frac{d y}{d x}\\ &\text { Multiplying by } y^{2} \text { to both the sides, we get }\\ &\Rightarrow \cos (x y)\left[x y^{2} \frac{d y}{d x}+y^{3}\right]+y-x \frac{d y}{d x}=2 x y^{2}-y^{2} \frac{d y}{d x}\\ &\Rightarrow x y^{2} \cos (x y) \frac{d y}{d x}+y^{3} \cos (x y)+y-x \frac{d y}{d x}=2 x y^{2}-y^{2} \frac{d y}{d x}\\ &\Rightarrow \mathrm{xy}^{2} \cos (\mathrm{xy}) \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{xy}^{2}-\mathrm{y}^{3} \cos (\mathrm{xy})-\mathrm{y}\\ &\Rightarrow \frac{d y}{d x}\left[x y^{2} \cos (x y)-x+y^{2}\right]=2 x y^{2}-y^{3} \cos (x y)-y\\ &\Rightarrow \frac{d y}{d x}=\frac{2 x y^{2}-y^{3} \cos (x y)-y}{x y^{2} \cos (x y)-x+y^{2}} \end{aligned}

Posted by

infoexpert22

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