Find (dy)/(dx) given: (a) xy^3-2x^2y^2+x^4=1 (b) x^2siny-ycosx=10x^3 xcosy-y^2sinx=2 (d) e^xy^(2)=10(x^2+y^2) ln(x^2+sqrt (y))=sin(xy^2)

To find $\frac{dy}{dx}$ for each equation, we will differentiate both sides of each equation with respect to $x$ using implicit differentiation.<br /><br />(a) $xy^{3}-2x^{2}y^{2}+x^{4}=1$<br /><br />Differentiating both sides with respect to $x$, we get:<br />$y^{3} + 3xy^{2}\frac{dy}{dx} - 4x^{2}y^{2}\frac{dy}{dx} - 4x^{3} = 0$<br /><br />Simplifying, we have:<br />$(3xy^{2} - 4x^{2}y^{2})\frac{dy}{dx} = -y^{3} + 4x^{3} - 1$<br /><br />Therefore, the derivative $\frac{dy}{dx}$ is:<br />$\frac{dy}{dx} = \frac{-y^{3} + 4x^{3} - 1}{3xy^{2} - 4x^{2}y^{2}}$<br /><br />(b) $x^{2}\sin y - y\cos x = 10x^{3}$<br /><br />Differentiating both sides with respect to $x$, we get:<br />$2x\sin y + x^{2}\cos y\frac{dy}{dx} - \cos x\frac{dy}{dx} = 30x^{2}$<br /><br />Simplifying, we have:<br />$(x^{2}\cos y - \cos x)\frac{dy}{dx} = 30x^{2} - 2x\sin y$<br /><br />Therefore, the derivative $\frac{dy}{dx}$ is:<br />$\frac{dy}{dx} = \frac{30x^{2} - 2x\sin y}{x^{2}\cos y - \cos x}$<br /><br />(c) $x\cos y - y^{2}\sin x = 2$<br /><br />Differentiating both sides with respect to $x$, we get:<br />$\cos y + x(-\sin y)\frac{dy}{dx} - 2y\sin x - y^{2}\cos x\frac{dy}{dx} = 0$<br /><br />Simplifying, we have:<br />$(x(-\sin y) - y^{2}\cos x)\frac{dy}{dx} = -\cos y - 2y\sin x$<br /><br />Therefore, the derivative $\frac{dy}{dx}$ is:<br />$\frac{dy}{dx} = \frac{-\cos y - 2y\sin x}{x(-\sin y) - y^{2}\cos x}$<br /><br />(d) $e^{xy^{2}} = 10(x^{2} + y^{2})$<br /><br />Differentiating both sides with respect to $x$, we get:<br />$y^{2}e^{xy^{2}} + 2xye^{xy^{2}}\frac{dy}{dx} = 20x$<br /><br />Simplifying, we have:<br />$y^{2}e^{xy^{2}} + 2xye^{xy^{2}}\frac{dy}{dx} = 20x$<br /><br />Therefore, the derivative $\frac{dy}{dx}$ is:<br />$\frac{dy}{dx} = \frac{20x - y^{2}e^{xy^{2}}}{2xye^{xy^{2}}}$<br /><br />(e) $\ln(x^{2} + \sqrt{y}) = \sin(xy^{2})$<br /><br />Differentiating both sides with respect to $x$, we get:<br />$\frac{2x}{x^{2} + \sqrt{y}} + \frac{1}{2\sqrt{y}}\frac{dy}{dx} = 2y^{2}\cos(xy^{2})$<br /><br />Simplifying, we have:<br />$\frac{1}{2\sqrt{y}}\frac{dy}{dx} = 2y^{2}\cos(xy^{2}) - \frac{2x}{x^{2} + \sqrt{y}}$<br /><br />Therefore, the derivative $\frac{dy}{dx}$ is:<br />$\frac{dy}{dx} = 2y^{2}\cos(xy^{2}) - \frac{2x}{x^{2} + \sqrt{y}} \cdot 2\sqrt{y}$