The directional derivative and gradient are very closely related. So we cover both on this page. First, a couple of quick definitions.
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Directional Derivative   You know from single variable calculus that the derivative is just the slope of the tangent line. If you are given a point, you can find the slope of the specific tangent line at that specific point. Similarly, the directional derivative is the slope of a tangent line but it applies to three dimensions where you are given a point and a unit vector. This unit vector tells you which specific tangent line you are talking about at that point. In summary, a directional derivative is the slope of a tangent line in a specific direction.
Gradient   The gradient is a vector (which is why its also called the gradient vector) calculated using the partial derivative. It contains information about the slope at every point. We can use the gradient to get the directional derivative.
Remember that in multivariable calculus most concepts use special notation. The directional derivative and gradient are no exception.
Here is a great video explaining the directional derivative and gradient in detail and the associated notation, including some examples. It is a long video, so we separate it into video clips on this page as the discussion continues.
video by Dr Chris Tisdell 

Gradient
\(\displaystyle{ \nabla g(x,y) = \frac{\partial g}{\partial x} \hat{i} + \frac{\partial g}{ \partial y}\hat{j} }\)
where \(\displaystyle{ \nabla }\) is the 'del' operator
\(\displaystyle{ \nabla = \frac{\partial }{\partial x}\hat{i} + \frac{\partial }{ \partial y}\hat{j} }\)
The gradient can also be indicated by grad(g) and can be called the grad of g. Notice that we do not use vector notation on \(\nabla g(x,y)\). We could write \( \vec{ \nabla } g(x,y)\) to clearly indicate that the result is a vector and some instructors do this. However, many of them don't and we won't either. So you just need to remember that \(\nabla g(x,y)\) is a vector (or, more precisely, a vector field).
Functions of More Than Two Variables
You can easily extend the definition of the gradient and directional derivative to functions of 3 or more variables. However, you lose the physical interpretation of slope. For three dimensions, the gradient looks like you would expect.
\(\displaystyle{ \nabla h(x,y,z) = \frac{\partial h}{\partial x} \hat{i} + \frac{\partial h}{ \partial y}\hat{j} + \frac{\partial h}{ \partial z}\hat{k} }\) 
Before we go on, let's make sure we understand this by working a few practice problems.
Practice  Gradient
Unless otherwise instructed, find the gradient of these functions.
\( f(x,y) = x\sin y + \cos y \)
Problem Statement 

Find the gradient of \( f(x,y) = x\sin y + \cos y \)
Final Answer 

\(\nabla f(x,y) = (\sin y) \vhat{i} + (x\cos y  \sin y)\vhat{j}\)
Problem Statement
Find the gradient of \( f(x,y) = x\sin y + \cos y \)
Solution
\(\displaystyle{ \nabla f(x,y) = \frac{\partial f}{\partial x} \vhat{i} + \frac{\partial f}{\partial y} \vhat{j} }\) 
\( = (\sin y + 0) \vhat{i} + (x\cos y  \sin y)\vhat{j} \) 
\( = (\sin y) \vhat{i} + (x\cos y  \sin y) \vhat{j} \) 
Final Answer
\(\nabla f(x,y) = (\sin y) \vhat{i} + (x\cos y  \sin y)\vhat{j}\)
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If \( f(x,y,z) = x^3+y+z^2 \), calculate \( \nabla f \) and \( \nabla f(1,2,3) \).
Problem Statement
If \( f(x,y,z) = x^3+y+z^2 \), calculate \( \nabla f \) and \( \nabla f(1,2,3) \).
Solution
video by Dr Chris Tisdell 

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Calculate \(\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa\tau^3} \right) }\) at \( (\tau, \kappa) = (1,5) \)
Problem Statement 

Calculate \(\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa\tau^3} \right) }\) at \( (\tau, \kappa) = (1,5) \)
Hint 

This problem uses unusual variables \( \tau \) and \( \kappa \). However, they work just as if you had t and x.
Problem Statement
Calculate \(\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa\tau^3} \right) }\) at \( (\tau, \kappa) = (1,5) \)
Hint
This problem uses unusual variables \( \tau \) and \( \kappa \). However, they work just as if you had t and x.
Solution
video by World Wide Center of Mathematics 

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Calculate the gradient vector field of \( f(x,y) = \ln(2x+5y) \).
Problem Statement
Calculate the gradient vector field of \( f(x,y) = \ln(2x+5y) \).
Solution
video by MIP4U 

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Given \( f(x,y) = xy \), determine \( \nabla f(3,2) \).
Problem Statement
Given \( f(x,y) = xy \), determine \( \nabla f(3,2) \).
Solution
video by MIP4U 

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\( f(x,y,z) = z e^{xy} \)
Problem Statement 

Find the gradient of \( f(x,y,z) = z e^{xy} \)
Final Answer 

\( \nabla f(x,y,z) = yze^{xy} \vhat{i}  xz e^{xy} \vhat{j} + e^{xy} \vhat{k} \)
Problem Statement
Find the gradient of \( f(x,y,z) = z e^{xy} \)
Solution
\(\displaystyle{ \nabla f(x,y,z) = \frac{\partial f}{\partial x} \vhat{i} + \frac{\partial f}{\partial y} \vhat{j} + \frac{\partial f}{\partial z} \vhat{k} }\) 
\(\displaystyle{ z(y)e^{xy} \vhat{i} + z (x) e^{xy} \vhat{j} + e^{xy} \vhat{k} }\) 
Final Answer
\( \nabla f(x,y,z) = yze^{xy} \vhat{i}  xz e^{xy} \vhat{j} + e^{xy} \vhat{k} \)
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Directional Derivative
\(\displaystyle{ D_{\vec{u}}f = \nabla f \cdot \vec{u} }\)
We read this as the directional derivative of f in the direction of \(\vec{u}\).
The directional derivative can be thought of as the magnitude of the gradient at a certain point in a specific direction. To calculate the directional derivative, we take the dot product of the gradient with the unit vector ( \(\vec{u}\) ) in the direction we are interested in. The equation looks like this.
Here is a video going through a proof showing that the directional derivative is the dot product of the gradient with a direction vector. This video is not required for understanding how to use the dot product version of the directional derivative but it will help you understand it. So it is a good video to watch.
video by Dr Chris Tisdell 

Here is a video clip discussing the properties of the directional derivative based on the use of the dot product.
video by Dr Chris Tisdell 

From the last video clip, you now know that the direction of maximum increase of a function f is the gradient \(\nabla f\) and the direction of maximum decrease is \(\nabla f\). ( Remember that these are vectors. )
Since the gradient is just a partial derivative, the basic rules of algebra are followed as well as the chain rule. If you would like to confirm these rules, here is a quick video clip to watch.
video by Dr Chris Tisdell 

Practice  Directional Derivative
Unless otherwise instructed, find the directional derivative of these functions.
Calculate the derivative of \( f(x,y) = xe^y + \cos(xy) \) at the point \( (2,0) \) in the direction of \( \vec{v} = 3\hat{i}  4\hat{j} \).
Problem Statement
Calculate the derivative of \( f(x,y) = xe^y + \cos(xy) \) at the point \( (2,0) \) in the direction of \( \vec{v} = 3\hat{i}  4\hat{j} \).
Solution
video by Dr Chris Tisdell 

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At the point \( (1,1) \), determine the directions in which \( f(x,y) = x^2/2 + y^2/2 \) (a) increases most rapidly; (b) decreases most rapidly; (c) has zero change
Problem Statement
At the point \( (1,1) \), determine the directions in which \( f(x,y) = x^2/2 + y^2/2 \) (a) increases most rapidly; (b) decreases most rapidly; (c) has zero change
Solution
video by Dr Chris Tisdell 

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Calculate the derivative of \( f(x,y,z) = x^3  xy^2z \) at the point \( (1,1,0) \) in the direction \( \vec{v} = 2\hat{i}  3\hat{j} + 6\hat{k}\).
Problem Statement
Calculate the derivative of \( f(x,y,z) = x^3  xy^2z \) at the point \( (1,1,0) \) in the direction \( \vec{v} = 2\hat{i}  3\hat{j} + 6\hat{k}\).
Solution
video by Dr Chris Tisdell 

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Calculate the directional derivative of \( f(x,y) = 1  x^2/2  y^4/4 \) in the direction \(\hat{i} + \hat{j}\) at the point \( (1,2) \).
Problem Statement
Calculate the directional derivative of \( f(x,y) = 1  x^2/2  y^4/4 \) in the direction \(\hat{i} + \hat{j}\) at the point \( (1,2) \).
Solution
video by Dr Chris Tisdell 

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Calculate the directional derivative of \( f(x,y) = x^2 + 2xy + 3y^2 \) at the point \( (2,1) \) in the direction \(\hat{i}+\hat{j}\).
Problem Statement
Calculate the directional derivative of \( f(x,y) = x^2 + 2xy + 3y^2 \) at the point \( (2,1) \) in the direction \(\hat{i}+\hat{j}\).
Solution
video by Krista King Math 

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Calculate the maximum directional derivative and its direction for \( f(x,y) = 2x^2 + 3xy + 4y^2 \) at \( (1,1) \).
Problem Statement
Calculate the maximum directional derivative and its direction for \( f(x,y) = 2x^2 + 3xy + 4y^2 \) at \( (1,1) \).
Solution
video by Krista King Math 

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Calculate the directional derivative of \( f(x,y) = x^2y^3  y^4 \) at \( (2,1) \) in the direction given by the angle \( \theta = \pi/4 \).
Problem Statement
Calculate the directional derivative of \( f(x,y) = x^2y^3  y^4 \) at \( (2,1) \) in the direction given by the angle \( \theta = \pi/4 \).
Solution
video by PatrickJMT 

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Calculate the value of the maximum rate of change of the function \( f(x,y) = \sin(xy) \) at the point \( (1,0) \).
Problem Statement
Calculate the value of the maximum rate of change of the function \( f(x,y) = \sin(xy) \) at the point \( (1,0) \).
Solution
video by Krista King Math 

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Calculate the direction of the maximum rate of change of the scalar field \( \phi(x,y,z) = 2xy + x\ln(y+z) \) at \( P(1,3,2) \) and the maximum value of the directional derivative at P.
Problem Statement
Calculate the direction of the maximum rate of change of the scalar field \( \phi(x,y,z) = 2xy + x\ln(y+z) \) at \( P(1,3,2) \) and the maximum value of the directional derivative at P.
Solution
video by Dr Chris Tisdell 

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Really UNDERSTAND Calculus
Topics You Need To Understand For This Page 

Links You May Find Helpful 
Better Explained: Vector Calculus: Understanding the Gradient 
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Practice Instructions
Unless otherwise instructed, find the gradient or directional derivative of these functions.