To differentiate $e^{x^2 + y^2 - xy}$, we need to apply the chain rule. The expression is a composite function where $u = x^2 + y^2 - xy$ is the exponent of $e^u$. We'll compute the partial derivatives with respect to both $x$ and $y$.

1. Derivative with respect to $x$:

$\frac{\partial}{\partial x} \left( e^{x^2 + y^2 - xy} \right) = e^{x^2 + y^2 - xy} \cdot \frac{\partial}{\partial x} (x^2 + y^2 - xy)$

Now, compute the partial derivative of $u = x^2 + y^2 - xy$ with respect to $x$:

$\frac{\partial}{\partial x} (x^2 + y^2 - xy) = 2x - y$

Thus, the partial derivative with respect to $x$ is:

$\frac{\partial}{\partial x} \left( e^{x^2 + y^2 - xy} \right) = e^{x^2 + y^2 - xy} \cdot (2x - y)$

2. Derivative with respect to $y$:

$\frac{\partial}{\partial y} \left( e^{x^2 + y^2 - xy} \right) = e^{x^2 + y^2 - xy} \cdot \frac{\partial}{\partial y} (x^2 + y^2 - xy)$

Now, compute the partial derivative of $u = x^2 + y^2 - xy$ with respect to $y$:

$\frac{\partial}{\partial y} (x^2 + y^2 - xy) = 2y - x$

Thus, the partial derivative with respect to $y$ is:

$\frac{\partial}{\partial y} \left( e^{x^2 + y^2 - xy} \right) = e^{x^2 + y^2 - xy} \cdot (2y - x)$

Conclusion:

The partial derivatives are:

• With respect to $x$: $e^{x^2 + y^2 - xy} \cdot (2x - y)$
• With respect to $y$: $e^{x^2 + y^2 - xy} \cdot (2y - x)$

Would you like further details or have any questions about this process?

Related questions:

1. How do you apply the chain rule in multivariable calculus?
2. What is the derivative of $e^{x^2}$ with respect to $x$?
3. How do you find the total derivative of a function involving both $x$ and $y$?
4. What is the difference between partial and total derivatives?
5. Can you explain implicit differentiation with respect to $y$ in a multivariable function?

Tip:

For exponential functions involving composite expressions, always remember to use the chain rule, which includes differentiating the exponent first and then multiplying by the derivative of the exponent.