# How can i determine the analytical expression of the derived funtion "df(x)/dx" ?

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Mallouli Marwa on 23 Jan 2017
Commented: Star Strider on 23 Jan 2017
How can i determine the analytical expression of the derived funtion "df(x)/dx"?
x=linspace (0:40)
L=3;
sol(1)=2;
psi=1;
f (x) = @(x) cos ( (sol(1)/L) *x)- cosh( (sol(1)/L) *x)+ psi *( sin( (sol(1)/L) *x)- sinh( (sol(1))

Star Strider on 23 Jan 2017
Use the Symbolic Math Toolbox:
syms x
L=sym(3);
sol(1)=sym(2);
psi=sym(1);
f (x) = cos ( (sol(1)/L) *x)- cosh( (sol(1)/L) *x)+ psi *( sin( (sol(1)/L) *x)- sinh( (sol(1))));
df_dx = diff(f,x)
df_dx_fcn = matlabFunction(df_dx)
df_dx(x) =
(2*cos((2*x)/3))/3 - (2*sin((2*x)/3))/3 - (2*sinh((2*x)/3))/3
df_dx_fcn = @(x) cos(x.*(2.0./3.0)).*(2.0./3.0)-sin(x.*(2.0./3.0)).*(2.0./3.0)-sinh(x.*(2.0./3.0)).*(2.0./3.0)
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Star Strider on 23 Jan 2017
I am not certain how you called the function, that is named ‘df_dx’, not ‘df_dy’. You can always rename it in the derivation and assignment, but you have to use the function name that exists. (I do not know if you used double quotes in your code. The double quotes (") are not valid MATLAB syntax.)
Depending on what you want, one of these will work:
y = 40;
SymOutput = df_dx(y)
SymOutput = vpa(df_dx(y))
NumOutput = df_dx_fcn(y)
SymOutput =
(2*cos(80/3))/3 - (2*sin(80/3))/3 - (2*sinh(80/3))/3
SymOutput =
-127076407709.60347598489525821488
NumOutput =
-127.0764e+009

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