how to find slope of sin^-1(x) curve at a given point x=sqrt(1/2)? the code of forward differentiation doesn't work for me

h=0.01

x = -1:h:1;

y=asind(x);

for i=1:length(x)-1

dy(i)=(y(x(i+1))-y(x(i)))/h;

end

Answer 1

[KSSV]

h=0.01 ;
x = -1:h:1;
y=asind(x);
dy = zeros(length(x)-1,1) ;
for i=1:length(x)-1
    dy(i)=(y(i+1)-y(i))/h;
end

OR

h=0.01 ;
x = -1:h:1;
y = @(x) asind(x);
dy = zeros(length(x)-1,1) ;
for i=1:length(x)-1
    dy(i)=(y(x(i+1))-y(x(i)))/h;
end

Actually no loop is needed:

h=0.01 ;
x = -1:h:1;
y=asind(x);
dy=diff(y)/h ; 

Answer 2

[David Hill]

dydx=@(x)-cos(x)./(sin(x).^2);
dydx(sqrt(1/2))