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This example shows how to analytically find and evaluate derivatives using Symbolic Math Toolbox™. In the example you will find the 1st and 2nd derivative of f(x) and use these derivatives to find local maxima, minima and inflection points.

Computing the first derivative of an expression helps you find local minima and maxima of that expression. Before creating a symbolic expression, declare symbolic variables:

`syms x`

By default, solutions that include imaginary components are included in the results. Here, consider only real values of `x`

by setting the assumption that `x`

is real:

`assume(x, 'real')`

As an example, create a rational expression (i.e., a fraction where the numerator and denominator are polynomial expressions).

f = (3*x^3 + 17*x^2 + 6*x + 1)/(2*x^3 - x + 3)

f =

Plotting this expression shows that the expression has horizontal and vertical asymptotes, a local minimum between -1 and 0, and a local maximum between 1 and 2:

fplot(f) grid

To find the horizontal asymptote, compute the limits of `f`

for `x`

approaching positive and negative infinities. The horizontal asymptote is `y = 3/2`

:

lim_left = limit(f, x, -inf)

lim_left =

lim_right = limit(f, x, inf)

lim_right =

Add this horizontal asymptote to the plot:

hold on plot(xlim, [lim_right lim_right], 'LineStyle', '-.', 'Color', [0.25 0.25 0.25])

To find the vertical asymptote of `f`

, find the poles of `f`

:

pole_pos = poles(f, x)

pole_pos =

Approximate the exact solution numerically by using the `double`

function:

double(pole_pos)

ans = -1.2896

Now find the local minimum and maximum of `f`

. If a point is a local extremum (either minimum or maximum), the first derivative of the expression at that point is equal to zero. Compute the derivative of `f`

using `diff`

:

g = diff(f, x)

g =

To find the local extrema of `f`

, solve the equation `g == 0`

:

g0 = solve(g, x)

g0 =

Approximate the exact solution numerically by using the `double`

function:

double(g0)

`ans = `*2×1*
-0.1892
1.2860

The expression `f`

has a local maximum at `x = 1.286`

and a local minimum at `x = -0.189`

. Obtain the function values at these points using `subs`

:

f0 = subs(f,x,g0)

f0 =

Approximate the exact solution numerically by using the `double`

function on the variable `f0`

:

double(f0)

`ans = `*2×1*
0.1427
7.2410

Add point markers to the graph at the extrema:

`plot(g0, f0, 'ok')`

Computing the second derivative lets you find inflection points of the expression. The most efficient way to compute second or higher-order derivatives is to use the parameter that specifies the order of the derivative:

h = diff(f, x, 2)

h =

Now Simplify that result:

h = simplify(h)

h =

To find inflection points of `f`

, solve the equation `h = 0`

. Here, use the numeric solver `vpasolve`

to calculate floating-point approximations of the solutions:

h0 = vpasolve(h, x)

h0 =

The expression `f`

has two inflection points: `x = 1.865`

and `x = 0.579`

. Note that `vpasolve`

also returns complex solutions. Discard those:

h0(imag(h0)~=0) = []

h0 =

Add markers to the plot showing the inflection points:

plot(h0, subs(f,x,h0), '*k') hold off