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## Multivariable calculus

### Course: Multivariable calculus>Unit 2

Lesson 2: Gradient and directional derivatives

# Directional derivative, formal definition

Learn the limit definition of a directional derivative. This helps to clarify what it is really doing.  Created by Grant Sanderson.

## Want to join the conversation?

• Isn't the vector "v" supossed to be a unit vector? Because if you were taking a scalar multiple of the vector v, and then computing the directional derivative, then the value of the directional derivative would change. I'm aware Grant mentions that when you double the size of the vector "v", that should double the size of your derivative, but is that really always the case? It seems to me that it would double the value only if the function was linear. •  When you wish to interpret the directional derivative as a certain slope, namely the slope you get by intersecting the graph with a plane pointing in the direction of your vector, v does indeed have to be a unit vector.

However, the directional derivative has meaning beyond the notion of slope, and often you actually do want to account for the length of your vector. For example, check out the multivariable chain rule videos.
• How do you get from the formal definition of the directional derivative to the formula of the previous video, the dot product of the gradient with the directional vector? • Can somedy explain how does adding the two derivatives with coefficients of a vector result in the derivative along that vector? • I don't know if this helps, but the way I thought about it was that in single variable calculus you still are sort of multiplying your h in f(a+h) by a vector. It's just that that vector happens to be (1, 0) so the y direction isn't actually taken into account. Does that make sense? So when you do the same thing in multivariable calculus, you aren't multiplying by a single dimension unit vector, instead you're multiplying by a multi dimensional vector that may not have similar values of change in the x and y position. That's the way I think about it anyway, hope that helps.
• Near the end of the video, he discusses that changing the value with which you scale the vector changes the derivative; for example, doubling the vector causes the derivative to double. However, since you are taking the value of the vector as h approaches 0, and the vector "v" is multiplied by h, why does the change in magnitude but no change in direction change the derivative? • I think it is because you should always remember the denominator part is the tiny movement of the input value, which means it should always be h times the scalar of the vector. If you use a unit vector, then the bottom part is h*|v| = h*1 = h; however, if you use a non-unit vector, say twice the length of the unit vector, then the bottom part should be h*|2|. I think it would be easier to understand if Grant just gives the definition with h|v| at the bottom since h itself is not the tiny movement on the vector direction.
• shouldn't it be h(i-hat) in the numerator too? • At the inception of the video (), why does Grant say partial derivative with respect to x? Shouldn't it instead be with respect to a?
(1 vote) • can someone explain me use of Absolute function in denominator ?
(1 vote) • I don't know if my understanding of v has to be a unit vector is correct, please correct me if I misinterpret it. So I think the reason why it has to be a unit vector is because the dx part (denominator) is the tiny movement for the input, and so it should always be not just h but h times the scalar of the vector, for the existance of h is just to shrink the size of movement to approach zero. The denominator should always be h×|v|, but for the sake of simplicity, people make the vector a unit vector so that the bottom part becomes h×1= h.   