Directional Derivative Calculator with steps
Directional derivative calculator is used to find the gradient and directional derivative of the given function. It takes the points of x & y coordinates along with the points of the vector. It is a type of derivative calculator.
How to use this directional derivative calculator?
Follow the below steps to find the directional derivatives of the functions.
- Input the multivariable function.
- To enter the math keys, hit the keypad icon.
- Write the values of \(U_1\&U_2\).
- Enter the x & y coordinates.
- Click the calculate button.
- Press the show more button to view the steps.
- To recalculate hit the clear button.
What is the directional derivative?
In calculus, the directional derivative of a multivariable differentiable function along with a vector v at the given point x intuitively represents the instantaneous rate of change of function, moving through x with velocity specified by v.
The directional derivative of a scalar function f(x) along with the vector v is a function \(∇_v\:f\) defined by the limit.
\(∇_v\:f\left(x\right)=\lim _{h\to 0}\left(\frac{f\left(x+hv\right)-f\left(x\right)}{h}\right)\)
The directional derivative used various notations such as:
\(∇_v\:f\left(x\right),\:f_v'\left(x\right),\:\partial \:_vf\left(x\right),\:v.∇f\left(x\right),\:or\:v.\frac{\partial \:f\left(x\right)}{\partial \:x}\)
The formula of the directional derivative.
The directional derivative is the dot product of the gradient and the normalized vector.
\(∇_v\left(f\left(x\right)\right)=∇f\left(x\right).\:\frac{v}{\left|v\right|}\)
Rules of the directional derivative
Below are some rules of directional derivatives.
| Rule Name | Rules |
| Sum rule | \(∇_v\left(f\left(x\right)+g\left(x\right)\right)\:=∇_v\:f\left(x\right)+∇_v\:g\left(x\right)\) |
| Difference rule | \(∇_v\left(f\left(x\right)-g\left(x\right)\right)\:=∇_v\:f\left(x\right)-∇_v\:g\left(x\right)\) |
| Constant factor rule | \(∇_v\left(cf\left(x\right)\right)\:=c∇_v\:f\left(x\right)\) |
| Product rule | \(∇_v\left(f\left(x\right)\cdot g\left(x\right)\right)\:=g\left(x\right)∇_v\:f\left(x\right)+f\left(x\right)∇_v\:g\left(x\right)\) |
How to calculate the directional derivative?
Following is a solved example of a directional derivative.
Example
Find the directional derivative of \(e^x+3y\) at (x, y) = (3, 4) along with the vector u = (1, 2).
Solution
Step 1: Write the given function with the gradient notation.
\(∇\left(f\left(x,y\right)\right)=\frac{\partial f\left(x,y\right)}{\partial x},\frac{\partial f\left(x,y\right)}{\partial y}\)
\(∇\left(e^x+3y\right)=\frac{\partial }{\partial x}\left(e^x+3y\right),\frac{\partial \:}{\partial y}\left(e^x+3y\right)\)
Step 2: Take the partial derivative of the above function with respect to x & y.
\(\frac{\partial \:}{\partial x}\left(e^x+3y\right)=e^x\)
\(\frac{\partial \:}{\partial y}\left(e^x+3y\right)=3\)
Step 3: Put the given points of x & y.
\(∇\left(e^x+3y\right)|_{\left(x,y\right)=\left(3,4\right)}=\left(e^3,3\right)\)
Step 4: Find the length of the given vector and normalize the vector.
\(|u⃗\:|=\sqrt{1^2+2^2}=\sqrt{5}\)
To normalize divide (1, 2) component by \(\sqrt{5}\).
\(\left(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)\)
Step 5: Take the dot product of the gradient and the normalized vector.
\(D_{u⃗\:}\left(e^x+3y\right)|_{\left(3,4\right)}\:=\left(e^3,3\right).\left(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right)\)
\(D_{u⃗\:}\left(e^x+3y\right)|_{\left(3,4\right)}\:=\frac{e^3+6}{\sqrt{5}}\)