gradient

The gradient of a scalar function f with respect to the vector v is the vector of the first partial derivatives of f with respect to each element of v.

Find the gradient vector of f(x,y,z) with respect to vector [x,y,z]. The gradient is a vector with these components.

syms x y z
f(x,y,z) = 2*y*z*sin(x) + 3*x*sin(z)*cos(y);
v = [x,y,z];
gradient(f,v)

ans(x, y, z) = 
$$\left(\begin{array}{c}3\hspace{0.17em}\cos \left(y\right)\hspace{0.17em}\sin \left(z\right)+2\hspace{0.17em}y\hspace{0.17em}z\hspace{0.17em}\cos \left(x\right)\\ 2\hspace{0.17em}z\hspace{0.17em}\sin \left(x\right)-3\hspace{0.17em}x\hspace{0.17em}\sin \left(y\right)\hspace{0.17em}\sin \left(z\right)\\ 2\hspace{0.17em}y\hspace{0.17em}\sin \left(x\right)+3\hspace{0.17em}x\hspace{0.17em}\cos \left(y\right)\hspace{0.17em}\cos \left(z\right)\end{array}\right)$$

Find the gradient of a function f(x,y), and plot it as a quiver (velocity) plot.

Find the gradient vector of f(x,y) with respect to vector [x,y]. The gradient is vector g with these components.

syms x y
f = -(sin(x) + sin(y))^2;
v = [x y];
g = gradient(f,v)

g = 
$$\left(\begin{array}{c}-2\hspace{0.17em}\cos \left(x\right)\hspace{0.17em}\left(\sin \left(x\right)+\sin \left(y\right)\right)\\ -2\hspace{0.17em}\cos \left(y\right)\hspace{0.17em}\left(\sin \left(x\right)+\sin \left(y\right)\right)\end{array}\right)$$

Now plot the vector field defined by these components. MATLAB® provides the quiver plotting function for this task. The function does not accept symbolic arguments. First, replace symbolic variables in expressions for components of g with numeric values. Then use quiver.

[X,Y] = meshgrid(-1:.1:1,-1:.1:1);
G1 = subs(g(1),v,{X,Y});
G2 = subs(g(2),v,{X,Y});
quiver(X,Y,G1,G2)

Since R2021b

Use symbolic matrix variables to define a matrix multiplication that returns a scalar.

syms X Y [3 1] matrix
A = Y.'*X

A = $$Y^{\mathrm{T}}\hspace{0.17em}X$$

Find the gradient of the matrix multiplication with respect to $$X$$.

gX = gradient(A,X)

gX = $$Y$$

Find the gradient of the matrix multiplication with respect to $$Y$$.

gY = gradient(A,Y)

gY = $$X$$

Since R2021b

Find the gradient of the multivariable function

with respect to the vector $$x=[x_{1,1},x_{1,2},x_{1,3}]$$.

Use a symbolic matrix variable to express the function $$f$$ and its gradient in terms of the vector $$x$$.

syms x [1 3] matrix
f = sin(x)*sin(x).'

f = $$\sin \left(x\right)\hspace{0.17em}{\sin \left(x\right)}^{\mathrm{T}}$$

g = gradient(f,x)

g = $$2\hspace{0.17em}\left(\cos \left(x\right)\odot {\mathrm{I}}_3\right)\hspace{0.17em}{\sin \left(x\right)}^{\mathrm{T}}$$

To express the gradient in terms of the elements of $$x$$, convert the result to a vector of symbolic scalar variables using symmatrix2sym.

g = symmatrix2sym(g)

g = 
$$\left(\begin{array}{c}2\hspace{0.17em}\cos \left(x_{1,1}\right)\hspace{0.17em}\sin \left(x_{1,1}\right)\\ 2\hspace{0.17em}\cos \left(x_{1,2}\right)\hspace{0.17em}\sin \left(x_{1,2}\right)\\ 2\hspace{0.17em}\cos \left(x_{1,3}\right)\hspace{0.17em}\sin \left(x_{1,3}\right)\end{array}\right)$$

Alternatively, you can convert $$f$$ and $$x$$ to symbolic expressions of scalar variables and use them as inputs to the gradient function.

g = gradient(symmatrix2sym(f),symmatrix2sym(x))

g = 
$$\left(\begin{array}{c}2\hspace{0.17em}\cos \left(x_{1,1}\right)\hspace{0.17em}\sin \left(x_{1,1}\right)\\ 2\hspace{0.17em}\cos \left(x_{1,2}\right)\hspace{0.17em}\sin \left(x_{1,2}\right)\\ 2\hspace{0.17em}\cos \left(x_{1,3}\right)\hspace{0.17em}\sin \left(x_{1,3}\right)\end{array}\right)$$

Since R2023a

Create a 3-by-1 vector as a symbolic matrix variable $\mathit{X}$. Create a scalar field that is a function of $\mathit{X}$ as a symbolic matrix function $A\left(\mathit{X}\right)$, keeping the existing definition of $\mathit{X}$.

syms X [3 1] matrix
syms A(X) [1 1] matrix keepargs

Find the gradient of $A\left(\mathit{X}\right)$ with respect to $\mathit{X}$. The gradient function returns an unevaluated formula.

gradA = gradient(A,X)

gradA(X) = $${\nabla }_X\mathrm{ }A\left(X\right)$$

Show that the divergence of the gradient of $A\left(\mathit{X}\right)$ is equal to the Laplacian of $A\left(\mathit{X}\right)$, that is ${\nabla }_{\mathit{X}}\cdot {\nabla }_{\mathit{X}}\mathit{A}\left(\mathit{X}\right)={\Delta }_{\mathit{X}}\mathit{A}\left(\mathit{X}\right)$.

divOfGradA = divergence(gradA,X)

divOfGradA(X) = $${\Delta }_X\mathrm{ }A\left(X\right)$$

lapA = laplacian(A,X)

lapA(X) = $${\Delta }_X\mathrm{ }A\left(X\right)$$

Show that the curl of the gradient of $A\left(\mathit{X}\right)$ is zero, that is ${\nabla }_{\mathit{X}}\times {\nabla }_{\mathit{X}}\mathit{A}\left(\mathit{X}\right)=0_{3,1}$.

curlOfGradA = curl(gradA,X)

curlOfGradA(X) = $${\mathrm{0}}_{3,1}$$