Thread: Partial derivative with exponential & product rule

1. Partial derivative with exponential & product rule

So I've been trying to solve this partial derivative with respect to either x or y but my answer (using the product rule) is way off of what the solutions manual is.

The original function is $\displaystyle f(x,y) = (x^2 + 4y^2) * e^(1-x^2-y^2)$

The solutions manual states the partial derivative with respect to x is $\displaystyle 2xe^(1-x^2-y^2) * (1-x^2-4y^2)$ and the partial derivative with respect to y is $\displaystyle 2ye^(1-x^2-y^2) * (4-x^2-4y^2)$.

The partial derivative with respect to xx is $\displaystyle 2e^(1-x^2-y^2) * (1-5x^2+2x^4-4y^2+8x^2y^2)$ and respect to yy is $\displaystyle 2e^(1-x^2-Y^2) * (4-x^2-20y^2+8y^4+2x^2y^2)$ and with respect to xy is $\displaystyle -4xye^(1-x^2-y^2) * (5-x^2-4y^2)$.

Is there a special way to using the product rule (and quotient rule for that matter) as far as partial derivatives go outside of holding the variable constant that you're not taking the derivative with respect to?

2. When you treat it the same as always, the product rule holds in Partial Derivatives:

We have:

$\displaystyle z = f(x,y) = (x^2 + 4y^2)e^{(1 - x^2 - y^2)}$

We need to find:

$\displaystyle \frac{\partial z}{\partial x} = 2xe^{(1 - x^2 - y^2)} - 2xe^{(1 - x^2 - y^2)}(x^2 + 4y^2)$

Factor out a $\displaystyle 2xe^{(1 - x^2 - y^2)}$:

$\displaystyle \frac{\partial z}{\partial x} = 2xe^{(1 - x^2 - y^2)}(1 - (x^2 + 4y^2))$

$\displaystyle \frac{\partial z}{\partial x} = 2xe^{(1 - x^2 - y^2)}(1 - x^2 - 4y^2)$

And it's the same method for $\displaystyle \frac{\partial z}{\partial y}$.

Hope that helped.