Let #y=f(x)=sqrtx#. Then, #y=f(x)=x^(1/2)#.
Now, we can use the Standard Form #(1) : d/dx(x^n)=n*x^(n-1)#.
In our case, as n=1/2, we have,
The First Derivative of #y=f(x)=x^(1/2)# denoted by #dy/dx#, or, #f'(x)#, is given by,
#dy/dx=f'(x)=1/2*x^(1/2-1)=1/2*x^(-1/2)#
Now, the Second Derivative, denoted by, #(d^2y)/dx^2 or, f''(x)# is defined by,
#(d^2y)/dx^2=d/dx{dy/dx}, or, (f')'(x)#.
It simply means that to find the second derivative of a given fun. #f#, we have to differentiate #f'#, i.e., the (first) derivative of #f# again.
So, for #(d^2y)/dx^2#, we will find #d/dx{1/2*x^(-1/2)}#, by using the
Std. Form #(1)# and, the Working Rule#(2):d/dx{k*u}=k*du/dx#, where, #k# is a const., &, #u#, a fun. of #x#.
Hence, #(d^2y)/dx^2=d/dx{1/2*x^(-1/2)}=1/2{-1/2*x^(-1/2-1)}=-1/4*x^(-3/2)#.
Using radicals, #(d^2y)/dx^2=-1/(4(sqrtx)^3)#.
