Find fyy (x,y): Partial Derivative of x2y3 + x4y + xe2y

hthorne21
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Find fyy (x,y) where f(x,y) = x2y3 + x4y + xe2y
 
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This is the part where you show your attempt. Also use ^ to denote power, I assume x2y3 means x^2 y^3. If you know TeX you can write that as x^2y^3
 
Try this: replace every x with the constant a... now the equation should look like it is in one variable, y. So what you need now is to find the second derivative with respect to y. Does that make life easier?
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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