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Directional Derivative Examples

Directional derivative calculator 

Directional derivative calculator is used to find the directional derivative of a multivariable function. This directional differential calculator takes the dot product of the gradient of the function and normalized vectors.

It takes the points of the given vector along with the points of x and y coordinates.

How does this directional derivative calculator work?

A directional differentiation calculator is an easy-to-use tool. Follow the below steps to find the directional differentiation of the functions.

  • Input the multivariable function.
  • Enter the points of the given vector i.e., \(U_1\&U_2\).
  • Write the coordinates points of x and y.
  • Click the calculate button.
  • Press the show more button to view the steps.
  • To enter another function, hit the reset button.

What is the directional derivative?

The rate of change of a multivariable function f(x, y) or f(x, y, z) at the coordinate points of x, y, & z (p = \(\left(x_0,y_0\right)\) or p = \(\left(x_0,y_0,z_0\right)\)) in the direction of unit vector u = (s, t) is known as a directional derivative.

The directional derivative can be denoted by different notations such as:

\(∇_u\:f\left(x,y\right),\:f_u'\left(x,y\right),\:\partial _uf\left(x,y\right),\:v.∇f\left(x,y\right),\:or\:u.\frac{\partial f\left(x,y\right)}{\partial x}\)

A directional derivative is a form of derivative which tells about the changes in the function as you move along with some unit vector u.

The formula of a directional derivative

The directional derivative uses the gradient and the normalized vector to calculate the direction. The formula of the directional derivative is given below.

\(∇_u\left(f\left(x,y\right)\right)=∇f\left(x,y\right).\:\frac{u}{\left|u\right|}\)

How to calculate the directional derivative?

The following is an example of a directional derivative.

Example

Calculate the directional derivative of \(x^3+3e^y\) at \(\left(x_0,y_0\right)=\left(4,5\right)\) and the unit vector is u = (3, 4).

Solution

Step 1: First of all, find the gradient of the function by taking the partial differentiation.

\(∇\left(f\left(x,y\right)\right)=\frac{\partial \:f\left(x,y\right)}{\partial x},\frac{\partial \:f\left(x,y\right)}{\partial y}\)

\(∇\left(x^3+3e^y\right)=\frac{\partial \:}{\partial \:x}\left(x^3+3e^y\right),\frac{\partial \:\:}{\partial \:y}\left(x^3+3e^y\right)\)

\(\frac{\partial \:\:}{\partial x}\left(x^3+3e^y\right)=3x^2\)

\(\frac{\partial \:\:}{\partial y}\left(x^3+3e^y\right)=3e^y\)

\(∇\left(x^3+3e^y\right)=3x^2,3e^y\)

Step 2: Substitute the values of x & y coordinates.

\(∇\left(3x^2+3e^y\right)|_{\left(x_0,y_0\right)=\left(4,5\right)}=\left(3\left(4\right)^2,\:3e^5\right)\)

\(∇\left(3x^2+3e^y\right)|_{\left(x_0,y_0\right)=\left(4,5\right)}=\left(48,\:3e^5\right)\)

Step 3: Calculate the magnitude of the given vector.

\(|u⃗\:|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\)

Step 4: Divide each component of the unit vector by the magnitude to normalize the vector.

\(u⃗\:=\:\left(\frac{3}{5},\:\frac{4}{5}\right)\)

Step 5: To find the directional derivative, take the dot product of the gradient and the normalized vector.

\( D_{u⃗\:}\left(x^3+3e^y\right)|_{\left(4,\:5\right)}\:=\left(48,\:3e^y\right).\left(\frac{3}{5},\:\frac{4}{5}\right)\)

\( D_{u⃗\:}\left(x^3+3e^y\right)|_{\left(4,\:5\right)}\:=\frac{144+12e^5}{5}\)

\( D_{u⃗\:}\left(x^3+3e^y\right)|_{\left(4,\:5\right)}\:=384.9916\)

References

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