Multivariable calculus and vector analysis

Partial derivative examples

Example 1

Let $f\left(x,y\right)=y^3{x}^2$. Calculate $\frac{\partial f}{\partial x}\left(x,y\right)$.

Solution: To calculate $\frac{\partial f}{\partial x}\left(x,y\right)$, we simply view $y$ as being a fixed number and calculate the ordinary derivative with respect to $x$. The first time you do this, it might be easiest to set $y=b$, where $b$ is a constant, to remind you that you should treat $y$ as though it were number rather than a variable. Then, the partial derivative $\frac{\partial f}{\partial x}\left(x,y\right)$ is the same as the ordinary derivative of the function $g\left(x\right)=b^3{x}^2$. Using the rules for ordinary differentiation, we know that

$$\begin{array}{lll}\hfill \frac{\mathrm{d}g}{\mathrm{d}x}\left(x\right)=2b^{3}x.& & \hfill \end{array}$$

Now, we remember that $b=y$ and substitute $y$ back in to conclude that

$$\begin{array}{lll}\hfill \frac{\partial f}{\partial x}\left(x,y\right)=2y^{3}x.& & \hfill \end{array}$$

Example 2

For the same $f$, calculate $\frac{\partial f}{\partial y}\left(x,y\right)$.

Solution: This time, we’ll just calculate the derivative with respect to $y$ directly without replacing $x$ with a constant. We just have to remember to treat $x$ like a constant and use the rules for ordinary differentiation. We don’t touch the $x^2$ and only differentiate the $y^3$ factor to calculate that

$$\begin{array}{lll}\hfill \frac{\partial f}{\partial y}\left(x,y\right)=3x^2{y}^2.& & \hfill \end{array}$$

Example 3

For the same $f$, calculate $\frac{\partial f}{\partial x}\left(1,2\right)$.

Solution: From example 1, we know that $\frac{\partial f}{\partial x}\left(x,y\right)=2y^{3}x$. To evaluate this partial derivative at the point $\left(x,y\right)=\left(1,2\right)$, we just substitute the respective values for $x$ and $y$:

$$\begin{array}{lll}\hfill \frac{\partial f}{\partial x}\left(1,2\right)=2\left(2^3\right)\left(1\right)=16.& & \hfill \end{array}$$

Example 4

For

$$\begin{array}{lll}\hfill f\left(x_1,x_2,x_3,x_4\right)=3\frac{cos\left(x_1{x}_4\right)sin\left(x_2^5\right)}{e^{x_2}+\left(1+x_2^2\right)∕\left(x_1{x}_2{x}_4\right)}+5x_1{x}_3{x}_4& & \hfill \end{array}$$

calculate $\frac{\partial f}{\partial x_3}\left(a,b,c,d\right)$.

Solution: Although this initially looks hard, it’s really any easy problem. The ugly term does not depend on $x_3$, so in calculating partial derivative with respect to $x_3$, we treat it as a constant. The derivative of a constant is zero, so that term drops out. The derivative is just the derivative of the last term with respect to $x_3$, which is

$$\begin{array}{lll}\hfill \frac{\partial f}{\partial x_3}\left(x_1,x_2,x_3,x_4\right)=5x_1{x}_4& & \hfill \end{array}$$

Substituting in the values $\left(x_1,x_2,x_3,x_4\right)=\left(a,b,c,d\right)$, we obtain the final answer

$$\begin{array}{lll}\hfill \frac{\partial f}{\partial x_3}\left(a,b,c,d\right)=5ad.& & \hfill \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

2a. The partial derivative

• Introduction to partial derivatives
• Partial derivatives by limit definition
• Partial derivative examples

2b. The derivative and differentiability

2c. The chain rule, directional derivative and gradient

2d. Applications of differential calculus

3. Integral calculus

4. Vector calculus

5. The fundamental theorems of vector calculus

6. Review material