Partial Derivatives in Multivariable Calculus

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• 1. Definition of Partial Derivatives
• 2. Formula Used
• 3. Solved Examples
• 4. Conclusion

Partial Derivatives:
Partial derivatives are used when a function depends on two or more variables. A partial derivative represents the rate of change of the function with respect to one variable, while keeping the other variables constant.

If $f(x, y)$ is a function of two variables $x$ and $y$, the partial derivative of $f$ with respect to $x$ is denoted as: $\boxed{\frac{\partial f}{\partial x}}$

Similarly, the partial derivative of $f$ with respect to $y$ is denoted as: $\boxed{\frac{\partial f}{\partial y}}$

In general, for a function $f(x_1, x_2, ..., x_n)$, the partial derivative with respect to any variable $x_i$ is: $\boxed{\frac{\partial f}{\partial x_i}}$

Example 1: Partial Derivative of a Multivariable Polynomial

Differentiate $f(x, y) = 3x^2y + 2xy^2 - 5x + y$ with respect to $x$.

Solution:

We apply the partial derivative with respect to $x$, treating $y$ as a constant:

$\dfrac{\partial}{\partial x} \left( 3x^2y + 2xy^2 - 5x + y \right)$

Differentiating each term:

• $\frac{\partial}{\partial x} (3x^2y) = 6xy$
• $\frac{\partial}{\partial x} (2xy^2) = 2y^2$
• $\frac{\partial}{\partial x} (-5x) = -5$
• $\frac{\partial}{\partial x} (y) = 0$ (since $y$ is constant)

Thus, the partial derivative is: $\boxed{6xy + 2y^2 - 5}$

Example 2: Partial Derivative of an Exponential Function

Differentiate $f(x, y) = e^{xy}$ with respect to $y$.

Solution:

We apply the partial derivative with respect to $y$, treating $x$ as a constant:

$\dfrac{\partial}{\partial y} \left( e^{xy} \right)$

Using the chain rule:

$\dfrac{\partial}{\partial y} \left( e^{xy} \right) = e^{xy} \cdot \dfrac{\partial}{\partial y} (xy)$

Since $\dfrac{\partial}{\partial y} (xy) = x$, we get:

$\dfrac{\partial}{\partial y} \left( e^{xy} \right) = x e^{xy}$

Thus, the partial derivative is: $\boxed{x e^{xy}}$

Example 3: Partial Derivative of a Logarithmic Function

Differentiate $f(x, y) = \ln(x^2 + y^2)$ with respect to $x$.

Solution:

We apply the partial derivative with respect to $x$, treating $y$ as a constant:

$\dfrac{\partial}{\partial x} \left( \ln(x^2 + y^2) \right)$

Using the chain rule:

$\dfrac{\partial}{\partial x} \left( \ln(u) \right) = \dfrac{1}{u} \cdot \dfrac{\partial u}{\partial x}$

Here, $u = x^2 + y^2$, so $\dfrac{\partial u}{\partial x} = 2x$.

Thus:

$\dfrac{\partial}{\partial x} \left( \ln(x^2 + y^2) \right) = \dfrac{1}{x^2 + y^2} \cdot 2x$

So, the final answer is: $\boxed{\frac{2x}{x^2 + y^2}}$