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

Let's correct the differentiation process for both equations.<br /><br />(a) Given: $xy^{3} - 2x^{2}y^{2} + x^{4} = 1$<br /><br />To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$ using the product rule and chain rule where necessary.<br /><br />\[<br />\begin{aligned}<br />&\frac{d}{dx}(xy^{3}) - \frac{d}{dx}(2x^{2}y^{2}) + \frac{d}{dx}(x^{4}) = \frac{d}{dx}(1) \\<br />&y^{3} + x \cdot 3y^{2} \cdot \frac{dy}{dx} - 2x^{2} \cdot 2y \cdot \frac{dy}{dx} - 4x^{3} = 0 \\<br />&y^{3} + 3xy^{2} \frac{dy}{dx} - 4x^{2}y \frac{dy}{dx} - 4x^{3} = 0 \\<br />&y^{3} - 4x^{2}y \frac{dy}{dx} + 3xy^{2} \frac{dy}{dx} = 4x^{3} \\<br />&y^{3} + (3xy^{2} - 4x^{2}y) \frac{dy}{dx} = 4x^{3} \\<br />&\frac{dy}{dx} = \frac{4x^{3} - y^{3}}{3xy^{2} - 4x^{2}y}<br />\end{aligned}<br />\]<br /><br />So, the derivative $\frac{dy}{dx}$ for the first equation is:<br /><br />\[<br />\frac{dy}{dx} = \frac{4x^{3} - y^{3}}{^{2} - 4x^{2}y}<br />\]<br /><br />(b) Given: $x^{2} \sin y - y \cos x = 10x^{3}$<br /><br />To find $\frac{dy}{dx}$, we differentiate both sides of the equation with respect to $x$ using the product rule and chain rule where necessary.<br /><br />\[<br />\begin{aligned}<br />&\frac{d}{dx}(x^{2sin y) - \frac{d}{dx}(y \cos x) = \frac{d}{dx}(10x^{3}) \\<br />&2x \sin y + x^{2} \cos y \cdot \frac{dy}{dx} - \cos x \cdot \frac{dy}{dx} - y \sin x = 30x^{2} \\<br />&2x \sin y + (x^{2} \cos y - \cos x) \frac{dy}{dx} = 30x^{2} + y \sin x \\<br />&\frac{dy}{dx} = \frac{30x^{2} + y \sin x - 2x \sin y}{x^{2} \cos y - \cos x}<br />\end{aligned}<br />\]<br /><br />So, the derivative $\frac{dy}{dx}$ for the second equation is:<br /><br />\[<br />\frac{dy}{dx} = \frac{30x^{2} + y \sin x - 2x \sin y}{x^{2} \cos y - \cos x}<br />\]<br /><br />These are the correct derivatives for the given equations.