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

Let's go through each equation step by step to find \(\frac{dy}{dx}\).<br /><br />### (a) \(xy^3 - 2x^2y^2 + x^4 = 1\)<br /><br />To find \(\frac{dy}{dx}\), we'll differentiate both sides of the equation with respect to \(x\):<br /><br />\[ \frac{d}{dx} \left( xy^3 - 2x^2y^2 + x^4 \right) = \frac{d}{dx} (1) \]<br /><br />Using the product rule and chain rule, we get:<br /><br />\[ y^3 + 3xy^2 \frac{dy}{dx} - 4xy^2 - 4x^2y \frac{dy}{dx} + 4x^3 = 0 \]<br /><br />Now, collect the terms involving \(\frac{dy}{dx}\):<br /><br />\[ (3xy^2 - 4x^2y) \frac{dy}{dx} = -y^3 + 4xy^2 - 4x^3 \]<br /><br />Finally, solve for \(\frac{dy}{dx}\):<br /><br />\[ \frac{dy}{dx} = \frac{-y^3 + 4xy^2 - 4x^3}{3xy^2 - 4x^2y} \]<br /><br />### (b) \(x^2 \sin y - y \cos x = 10x^3\)<br /><br />Differentiate both sides with respect to \(x\):<br /><br />\[ \frac{d}{dx} \left( x^2 \sin y - y \cos x \right) = \frac{d}{dx} (10x^3) \]<br /><br />Using the product rule and chain rule, we get:<br /><br />\[ 2x \sin y + x^2 \cos y \frac{dy}{dx} + y \sin x \frac{dy}{dx} = 30x^2 \]<br /><br />Collect the terms involving \(\frac{dy}{dx}\):<br /><br />\[ (x^2 \cos y + y \sin x) \frac{dy}{dx} = 30x^2 - 2x \sin y \]<br /><br />Finally, solve for \(\frac{dy}{dx}\):<br /><br />\[ \frac{dy}{dx} = \frac{30x^2 - 2x \sin y}{x^2 \cos y + y \sin x} \]<br /><br />### (c) \(x \cos y - y^2 \sin x = 2\)<br /><br />Differentiate both sides with respect to \(x\):<br /><br />\[ \frac{d}{dx} \left( x \cos y - y^2 \sin x \right) = \frac{d}{dx} (2) \]<br /><br />Using the product rule and chain rule, we get:<br /><br />\[ \cos y - x \sin y \frac{dy}{dx} - 2y \cos x \frac{dy}{dx} = 0 \]<br /><br />Collect the terms involving \(\frac{dy}{dx}\):<br /><br />\[ (\cos y - 2y \cos x) \frac{dy}{dx} = -\cos y \]<br /><br />Finally, solve for \(\frac{dy}{dx}\):<br /><br />\[ \frac{dy}{dx} = \frac{-\cos y}{\cos y - 2y \cos x} \]<br /><br />### (d) \(e^{xy^2} = 10(x^2 + y^2)\)<br /><br />Differentiate both sides with respect to \(x\):<br /><br />\[ \frac{d}{dx} \left( e^{xy^2} \right) = \frac{d}{dx} \left( 10(x^2 + y^2) \right) \]<br /><br />Using the chain rule, we get:<br /><br />\[ e^{xy^2} \left( y^2 + 2xy \frac{dy}{dx} \right) = 20x + 20y \frac{dy}{dx} \]<br /><br />Collect the terms involving \(\frac{dy}{dx}\):<br /><br />\[ e^{xy^2} y^2 + 2xy e^{xy^2} \frac{dy}{dx} = 20x + 20y \frac{dy}{dx} \]<br /><br />Finally, solve for \(\frac{dy}{dx}\):<br /><br />\[ \frac{dy}{dx} = \frac{20x - e^{xy^2} y^2}{2xy e^{xy^2} - 20y} \]<br /><br />These are the derivatives \(\frac{dy}{dx}\) for each of the given equations.