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).
