MATLAB: How to get function value

derivativefunctionsymbolicSymbolic Math Toolboxsymsvalue

Hi I want to evaluate function and its derivative at a=2 , how to do that?

syms a
b=2
e=0.9 
q=1.2
g=2
k=3
d=0.5
f=(q*b*e/a-d*e)/((q^2)*(e^2)/a+2*e*g)
diff(f)

Best Answer

df(a) = diff(f)
df(2)

Related Solutions

MATLAB: What does “z” stands for in “solve” function

root(2*G^2*Q^2*m*z^3 - G^2*Q^2*m^2*z^3 - G^2*Q^2*z^3 - 4*G^2*Q^2*m*z^2 + 2*G^2*Q^2*m^2*z^2 + 2*G^2*Q^2*z^2 - G^2*m^2*z^2 - 2*m*z^2 + m^2*z^2 + z^2 + 2*G^2*Q^2*m*z - G^2*Q^2*m^2*z + 2*G^2*m*z - G^2*Q^2*z - G^2, z, 1)

stands for the "first" (#1) value, z, such that 2*G^2*Q^2*m*z^3 - G^2*Q^2*m^2*z^3 - G^2*Q^2*z^3 - 4*G^2*Q^2*m*z^2 + 2*G^2*Q^2*m^2*z^2 + 2*G^2*Q^2*z^2 - G^2*m^2*z^2 - 2*m*z^2 + m^2*z^2 + z^2 + 2*G^2*Q^2*m*z - G^2*Q^2*m^2*z + 2*G^2*m*z - G^2*Q^2*z - G^2 is 0.

Notice that expression is a cubic in z, so there are up to three solutions. Exactly what is meant by the "first" root(EXPRESSION,z,1) , or "second" root(EXPRESSION,z,2) or "third" root(EXPRESSION,z,3) is not really defined for user purposes. Whatever the indexing means, it is used consistently by MATLAB: if you were to fill in all of the variables with numerics, then #1 would always refer to the same root for those coefficients. On the other hand, if you start changing the coefficients, the identities of the roots can appear to change places from our perspective -- if you were to plot changing a parameter but using a constant root number, then you might see an apparent discontinuity, which is not an actual discontinuity in the roots and is only a discontinuity in whatever scheme is used internally to number the roots.

If you were to change your code to

x = solve(eqn, f, 'MaxDegree', 3)

then solve() would construct explicit equations instead of using root() placeholders. This can take some time. This can be used for up to 'MaxDegree', 4 -- that is, you can ask for explicit solutions up to quartics. It isn't usually a good idea, though.

MATLAB: What’s wrong with the folow expression

diff(f) would invoke f with no arguments, fetch its return value, and apply the numeric diff() function to it, where the numeric diff() function is (X(2:end) - X(1:end-1)).

If you have the symbolic toolbox, you could try

syms t
df = diff(f(t),t);
f2 = diff(df,t);
f1 = inline(char(f(t)));
df1 = inline(char(df));
df2 = inline(char(f2));

Or more directly,

syms t
ft = f(t);
df = diff(ft,t);
f2 = diff(df, t);
f1 = @f;
df1 = matlabFunction(df, 'vars', t);
df2 = matlabFunction(f2, 'vars', t);
f

Best Answer

Related Solutions

MATLAB: What does “z” stands for in “solve” function

MATLAB: What’s wrong with the folow expression

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