# 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 $f(x,y) = (x^2 + 4y^2) * e^(1-x^2-y^2)$

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

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

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:

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

We need to find:

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

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

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

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

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

Hope that helped.