Partial derivatives are one of the most basic concepts in mathematics, especially multivariable calculus and are widely used in physics, engineering and economics among other fields. Partial derivatives are important for students as many topics such as gradient vectors, optimizations, and differential equations are based on it.

The scope of the following article is to give the reader a general idea of what partial derivatives are and review several exercises to solidify the concept.

## What are Partial Derivatives?

Partial derivatives is a mathematical concept used in vector calculus and differential geometry. The term ‘partial’ indicates that if the function is dependent on more than one variable, then the derivative will be taken considering one variable to calculate the change concerning the chosen variable. The partial derivative of a function f(x, y) is represented by ∂f/∂x. Partial derivatives show how the value of a function changes as one of its variables changes while others stay the same.

### Notation

Partial Derivative of f with respect to x**∂f/∂x:****f**_{x}Other common notation for the partial derivative of f with respect to x**:**

## Partial Derivatives Practice Problems

**Example 1: Find the partial derivatives ∂f/∂x and ∂f/∂y for f(x, y) = 3x**^{2}** y + 4y**^{3}**.**

**Solution:**

f(x, y) = 3x

^{2}y + 4y^{3}∂f/∂x = ∂/∂x (3x

^{2}y + 4y^{3})= 3 . 2xy + 0 = 6xy

∂f/∂x = ∂/∂y (3x

^{2}y + 4y^{3})= 3x

^{2}+ 4 . 3y^{3}= 3x^{2}+ 12y^{2}

**Example 2: Given f(x,y) = ln(xy), compute ∂f/∂x and ∂f/∂y.**

**Solution:**

f(x, y) = ln(xy)

Using the chain rule:

∂f/∂x = ∂/∂x ln(xy) =

1/xy . y

= y/xy = 1/x

∂f/∂y = ∂/∂y ln(xy)

= 1/xy . x

= x/xy = 1/y

**Example 3: Determine the second order partial derivatives ∂**^{2}**f/∂x**^{2}**, ∂f**^{2}**/∂y**^{2}**, and ∂**^{2}**f/∂x∂y for f(x, y) = x**^{2}**e**^{y}**.**

**Solution:**

f(x, y) = x

^{2}e^{y}First, find the first order partial derivatives:

∂f/∂x = ∂/∂x(x

^{2}e^{y}) = 2xe^{y}∂f/∂x = ∂/∂y(x

^{2}e^{y}) = x^{2}e^{y}Now, find second order partial derivatives:

∂

^{2}/∂x = ∂/∂x(2xe^{y}) = 2e^{y}∂

^{2}/∂y^{2}= ∂/∂y(x^{2}e^{y}) = x^{2}e^{y}∂

^{2}f/∂x∂y = ∂/∂y(2xe^{y}) = 2xe^{y}

**Example 4: For f(x, y) = x**^{3 }**+ y**^{3}** – 3xy, find all the first and second order partial derivatives.**

**Solution:**

f(x, y) = x

^{3 }+ y^{3}– 3xyFirst order partial derivatives:

∂f/∂x = ∂/∂x(x

^{3 }+ y^{3 }– 3xy)= 3x

^{2 }– 3y∂f/∂x = ∂/∂y(x

^{3}+ y^{3}− 3xy)= 3y

^{2}– 3xSecond order partial derivatives:

∂

^{2}f/∂x^{2}= ∂/∂x(3x^{2 }− 3y) = 6x∂

^{2}f/∂y^{2 }= ∂/∂y(3y^{2}– 3x) = 6y∂

^{2}f/∂x∂y = ∂/∂y(3x^{2 }– 3y) = -3∂

^{2}f/∂y/∂x = ∂/∂x(3y^{2}– 3x) = -3

**Example 5: Find the equation of the tangent plane to the surface z= x**^{2}** + y**^{2}**at the point (1, 1, 2).**

**Solution:**

Function is f(x, y) = x

^{2 }+ y^{2}First, find the partial derivatives

∂f/∂x = 2x

∂f/∂x = 2y

At the point (1, 1, 2):

∂f/∂x (1, 1) = 2 ⋅ 1 = 2

∂f/∂x (1, 1) = 2 ⋅ 1 = 2

Equation of the tangent plane is:

z – z

_{0}= ∂f/∂x(x_{0}, y_{0})(x – x_{0}) + ∂f/∂x(x_{0}, y_{0})(y – y_{0})z − 2 = 2(x − 1) + 2(y − 1)

z − 2 = 2x − 2 + 2y − 2

z = 2x + 2y − 2

**Example 6: Given f(x, y) = xe**^{xy}**, find ∂f/∂x and ∂f/∂y.**

**Solution:**

Using the product rule:

f(x, y) = xe

^{xy}∂f/∂x = ∂/∂x(xe

^{xy})= e

^{xy}+ x . e^{xy}. y=e

^{xy }. y = e^{xy}(1 + xy)∂f/∂x = ∂/∂y(xe

^{xy})= x . e

^{xy }. x = x^{2}e^{xy}

**Example 7: For f(x, y) = sin(x + y), find ∂f/∂x and ∂f/∂y.**

**Solution:**

f(x, y) = sin(x + y)

∂f/∂x = ∂/∂x sin(x + y)

= cos(x + y) . ∂/∂x(x + y)

= cos(x + y) . 1 = cos(x + y)

∂f/∂x = ∂/∂y sin(x + y)

= cos(x + y) . ∂/∂y(x + y)

= cos(x + y) . 1 = cos(x + y)

**Example 8: Find the partial derivatives ∂f/∂x and ∂f/∂y f(x, y) = x**^{2 }**+xy + y**^{2}**.**

**Solution:**

f(x, y) = x

^{2 }+ xy + y^{2}∂f/∂x = ∂/∂x(x

^{2}+ xy + y)= 2x + y

∂f/∂x = ∂/∂y(x

^{2}+ xy + y^{2})= x + 2y

## Partial Derivatives Worksheet

**Q1. Find the partial derivatives ∂f/∂x and ∂f/∂y for f(x, y) = 3x**^{2}**y _ 4y**^{2}**.**

**Q2. Given f(x, y) = ln(x, y), compute ∂f/∂x and ∂f/∂y.**

**Q3. Determine the second order partial derivatives ∂**^{2}**f/∂x**^{2}**, ∂**^{2}**f/∂y**^{2}**, and ∂**^{2}**f/∂x∂y for f(x, y) = x**^{2}**e**^{y}**.**

**Q4. For f(x, y) = x**^{3 }**+ y3 −3xy, find all the first and second order partial derivatives.**

**Q5. Find all the 1st order derivatives of the given function f(u, v) = u**^{2}**sin(u+v**^{3}**) − sec(4u)tan**^{−1}**(2v**

**Q6. Find the equation of the tangent plane to the surface z = x**^{2}** + y**^{2}** at the point(1, 1, 2).**

**Q7. Locate and classify the critical points of f (x, y) = x**^{2}** − 4xy + 4y**^{2}**. **

**Q8. Find all the 1st order derivatives of the given function f(x, y, z) = 4x**^{3}**y**^{2}** − e**^{z}**y**^{4}** + z**^{3}**/ x**^{2}** + 4y − x**^{16 }** .**

**Q9. Letf(x, y)=(x-y)2. Determine the equations and shapes of the cross-sections whenx=0,y=0,x=y, and describe the level curves. Use a three-dimensional graphing tool to graph the surface. **

**Q10. If z = f(x, y) = x**^{4}**y**^{3 }**+ 8x**^{2}**y + y**^{4 }**+ 5x, then,**

**∂z/∂x =?****∂z/∂y =?**

## Conclusion

Partial derivatives are one of the core problems of the multivariable calculus theory and finding their realizations in different branches of knowledge. This way students will be able to appreciate it or the geometric and theoretical aspects behind it more and master the problems. In this article, the author gave an introduction, elaboration, and examples procedures of partial derivatives to help with understanding them.

**Also Check:**

- Algebra in Mathematics
- Algebra vs Calculus

## Frequently Asked Questions (FAQs)

### What is Difference between Partial Derivative and Ordinary Derivative?

The major difference between a partial derivative and an ordinary derivative is that a partial derivative means conventional differentiation where a certain function is considered with respect to one of the variables not simultaneously other variables.

### How are Partial Derivatives used in Optimization Problems?

First derivatives are used to find maximum or minimum points of the function, while the second order first derivatives are used to determine the type of critical points; whether they are maximum, minimum or saddle points.

### What is the Geometric Interpretation of Partial Derivatives?

Partial derivatives give the degree of inclination of the line tangent to the curve at a particular point, while keeping all other variables constant.

### Can Partial Derivatives be Higher Order?

Indeed, first-order partial derivatives can be of higher order if one differentiates the said first-order partial derivatives further.

### How do you find the Tangent Plane to a Surface at a Given Point?

Thus, the equation of this tangent plane is also determined with the help of the partial derivative of the function in the given point.

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