X
ADVERTISEMENT

Gradient Calculator

Enter the vector function and the values of the point in the gradient calculator. Click calculate to solve.

for( x, y )
for( x, y, z )
ADVERTISEMENT
ADVERTISEMENT

Is This Tool Helpful?

     

Gradient Calculator 

This gradient calculator finds the partial derivatives of functions. You can enter the values of a vector line passing from 2 points and 3 points. For detailed calculation, click “show steps”.

What is a Gradient?

The gradient is similar to the slope. It is represented by ∇(nabla symbol). A gradient in calculus and algebra can be defined as:

“A differential operator applied to a vector-valued function to yield a vector whose components are the partial derivatives of the function with respect to its variables.”

Gradient Formula

f(x,y,z)=[dfdx  dfdy  dfdz] \nabla f (x, y, z) = \Huge [ \small \begin{array}{ccc} \dfrac{df}{dx}\ \ \dfrac{df}{dy}\ \ \dfrac{df}{dz} \end{array} \Huge]

A gradient is calculated by finding the partial derivative of the function with respect to the variable. 

How to calculate gradient?

Example:

A function is x2 + y2. Find its gradient for point (2,2).

Solution:

Step 1: Find the differentiate.

 

f=(dfdx,dfdy)=ddx[d(x2+y2),=ddy[d(x2+y2) \nabla f = \Big( \dfrac{df}{dx}, \dfrac{df}{dy} \Big)=ddx[d(x2+ y2) , = ddy[d(x2+ y2)

 

=ddx[d(x2+y2),=ddy[d(x2+y2) =\dfrac{d}{dx}[d(x^2+ y^2) \enspace ,=\dfrac{d}{dy}[d(x^2+ y^2)

 

=d(ddx[x2]+ddx[y2]),=d(ddy[x2]+ddy[y2]) =d(\dfrac{d}{dx}[x^2]+\dfrac{d}{dx}[y^2]) \enspace, = d(\dfrac{d}{dy}[x^2]+\dfrac{d}{dy}[y^2])

 

=d(2x+0),=d(0+2y) = d (2x + 0) \enspace , = d (0 + 2y)

 

=2x,=2y = 2x , = 2y

 

Step 2: Plug in the point.

 

f(x,y)=[(2)(x),(2)(y)] \nabla f (x,y) = [(2)(x) , (2)(y)]

 

f(2,2)=[(2)(2),(2)(2)] \nabla f (2,2) = [(2)(2) , (2)(2)]

 

= 4 , 4

Step 3: Write equation.

 

(x2+y2)(x,y)=(2x,2y) \nabla(x^2+y^2)(x,y)=(2x , 2y)

 

(x2+y3)(x,y)=(2,2)=(4,4) \nabla(x^{2} + y^{3})|_{(x,y)=(2,2)} = (4,4)

 

AdBlocker Detected!

To calculate result you have to disable your ad blocker first.