How do you do a derivative with a certain value?

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?

[RASHID MALIK ANSARI]

I also have same problem

Answer 1

[Geoff Hayes]

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.

[Yianni]

Perfect thank you!

Answer 2

[John D'Errico]

help subs

[Kiana]

I dont see how this would help for the situation when the derivative is evaluated at an initial value for example as a condition to solve an ODE

[Torsten]

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]

@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