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

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)

2 个评论
显示无隐藏无

Mallouli Marwa 2017-1-23

If i want to comput the derived function in a point y=40 "df_dy(40)"

I obtain this error

Error using mupadmex

Error in MuPAD command: Index exceeds matrix dimensions.

Star Strider 2017-1-23

在 MATLAB Online 中打开

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