The function $x \sqrt{1}-x^{2},(x>0)$ has
1) A local maxima
2) A local minima
3) Neither local maxima nor a local minima
4) None of the above

Hint: Here we have to find whether the function is $x \sqrt{1}-x^{2},(x>0)$ a local maxima or a local minima. One of the most prevalent ideas in differential calculus is maxima and minima. The maxima and minima of the function are studied in a subfield of mathematics called "Calculus of Variations." The calculus of variations deals with functional variations when a slight modification to the function results in a modification to the functional value.

Formula Used:
If $f'(k)=0$ and $f''(k)=0$, then $x = k$ is a point of local maxima. The local maxima is the point at which $x=k$, and $f(k)$ is known as the local maximum value of $f(x)$.

Complete step by step Solution:
Given function is
$f(x)=x \sqrt{1}-x^{2}$
Differentiation is the ratio of a slight change in one quantity to a little change in another that depends on the first quantity. Calculus' main emphasis on the differentiation of a function makes it one of the subject's key ideas.
Differentiation is the process of determining the maximum or lowest value of a function, the speed and acceleration of moving objects, and the tangent of a curve.
$f^{\prime}(x)=\left[1-2 x^{2}\right] /\left[\sqrt{1}-x^{2}\right]=0$
$x=\pm 1 / \sqrt{2}$
But as $x>0$, we have $x=1 / \sqrt{2}$
$f^{\prime \prime}(x)=\left\{\left[\sqrt{1}-x^{2}\right](-4 x)-\left(1-2 x^{2}\right)\left(-x / \sqrt{1}-x^{2}\right)\right\} /\left(1-x^{2}\right)$
$=\left[2 x^{3}-3 x\right] /\left(1-x^{2}\right)^{3 / 2}$
$f^{\prime \prime}(1 / \sqrt{2})=\text { negative }$
Then $f(x)$ is maximum at $x=1 / \sqrt{2}$.

Hence, the correct option is 1.

Note:The first variation is the change in the function that is linear in nature, while the second variation is the change in the function that is quadratic in nature. The definite integrals that include the functions and their derivatives are how functional relationships are expressed.
The Euler-Lagrange method of the calculus of variations can be used to find the functions that maximize or minimize the function. The obvious meaning of the two Latin words maxima and minima is the highest and minimum values of a function, respectively. The term "Extrema" refers to both peaks and minima taken together. Here, we suppose that the domain of our function is continuous.