le answer above is NOT correct. ind dy/dx by implicit differentiation. e^x^(2y)=x+y dy/dx=

To find \(\frac{dy}{dx}\) by implicit differentiation for the equation \(e^{x^2 y} = x + y\), follow these steps:<br /><br />1. Differentiate both sides of the equation with respect to \(x\).<br /><br />Starting with the left side:<br />\[ \frac{d}{dx} \left( e^{x^2 y} \right) \]<br /><br />Using the chain rule:<br />\[ \frac{d}{dx} \left( e^{x^2 y} \right) = e^{x^2 y} \cdot \frac{d}{dx} (x^2 y) \]<br /><br />Now, apply the product rule to differentiate \(x^2 y\):<br />\[ \frac{d}{dx} (x^2 y) = 2x y + x^2 \frac{dy}{dx} \]<br /><br />So,<br />\[ \frac{d}{dx} \left( e^{x^2 y} \right) = e^{x^2 y} \cdot (2x y + x^2 \frac{dy}{dx}) \]<br /><br />Now, differentiate the right side:<br />\[ \frac{d}{dx} (x + y) = 1 + \frac{dy}{dx} \]<br /><br />Putting it all together:<br />\[ e^{x^2 y} \cdot (2x y + x^2 \frac{dy}{dx}) = 1 + \frac{dy}{dx} \]<br /><br />2. Solve for \(\frac{dy}{dx}\).<br /><br />First, distribute \(e^{x^2 y}\) on the left side:<br />\[ 2x y e^{x^2 y} + x^2 e^{x^2 y} \frac{dy}{dx} = 1 + \frac{dy}{dx} \]<br /><br />Next, isolate \(\frac{dy}{dx}\) terms on one side:<br />\[ x^2 e^{x^2 y} \frac{dy}{dx} - \frac{dy}{dx} = 1 - 2x y e^{x^2 y} \]<br /><br />Factor out \(\frac{dy}{dx}\) on the left side:<br />\[ \left( x^2 e^{x^2 y} - 1 \right) \frac{dy}{dx} = 1 - 2x y e^{x^2 y} \]<br /><br />Finally, solve for \(\frac{dy}{dx}\):<br />\[ \frac{dy}{dx} = \frac{1 - 2x y e^{x^2 y}}{x^2 e^{x^2 y} - 1} \]<br /><br />So, the derivative \(\frac{dy}{dx}\) is:<br />\[ \frac{dy}{dx} = \frac{1 - 2x y e^{x^2 y}}{x^2 e^{x^2 y} - 1} \]