How do you do a derivative with a certain value?

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Yianni
Yianni il 24 Ott 2014
Commentato: John D'Errico il 2 Dic 2022
My initial equation is f(x) = (3*x^5)-(x^3)+(2*x^2)+4 and when I derive that, I get f'(x)=15*x^4 - 3*x^2 + 4*x. How do I evaluate f'(x) when x = 1.7?
So far I have done this:
clear all, close all
syms x f = (3*x^5)-(x^3)+(2*x^2)+4; diff(f)
This gives me the derivative, but how do I find the value of f'(x) when I have a value for x?

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Geoff Hayes
Geoff Hayes il 24 Ott 2014
Yianni - try using subs as
syms x
f = (3*x^5)-(x^3)+(2*x^2)+4;
df = diff(f);
result = double(subs(df,1.7));
See also combine subs and double for numeric evaluations for more details.

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John D'Errico
John D'Errico il 24 Ott 2014
help subs
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Torsten
Torsten il 2 Dic 2022
Modificato: Torsten il 2 Dic 2022
Ran in:
In this case, the derivative is the ODE expression itself.
E.g.
syms x y
dydx = x + y; % ODE is dy/dx = x + y
x0 = 1;
y0 = 2; % initial condition is y(1) = 2
dydx0 = subs(dydx,[x y],[x0 y0]) % derivative at x0 = 1 is dydx0 = x0 + y0 = 3
dydx0 = 
3
John D'Errico
John D'Errico il 2 Dic 2022
Ran in:
@Kiana
That was not the question, to solve an ODE, or even work with one. The explicit question was asked as how to evaluate a function that you have differentiated. subs will do that.
But your question is simple to answer. Say your ODE is something easy:
y' = t^2 +2
Now evaluate the right hand side is easy.
syms y(t)
RHS = t^2 + 2;
Now you want to evaluate the right hand side for some value. Say, at t==1. You can use subs here
subs(RHS,t,1)
ans = 
3
or you can converty the relation into a MATLAB function, as:
RHSfun = matlabFunction(RHS)
RHSfun = function_handle with value:
@(t)t.^2+2.0
RHSfun(1)
ans = 3

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