Question

d/dx(x/2√a²-x² + a²/2 sin⁻¹ x/a)=......... (a > 0),Select Proper option from the given options.
(a) 1/√a²-x²
(b) √a²-x²
(c) √x²-a²
(d) √x²+a²

Answer 1

I guessed your question is $\frac{d}{dx}\{\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}\}$

so, $\frac{d}{dx}\{\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}\}$

=$\frac{d}{dx}\{\frac{x}{2}\sqrt{a^2-x^2}\}+\frac{d}{dx}\{\frac{a^2}{2}sin^{-1}\frac{x}{a}\}$

=$\frac{1}{2}[\sqrt{a^2-x^2}\frac{dx}{dx}+x\frac{d\sqrt{a^2-x^2}}{dx}]+\frac{a^2}{2}\frac{1}{\sqrt{1-\frac{x^2}{a^2}}}\frac{d\{\frac{x}{a}\}}{dx}$

=$\frac{1}{2}[\sqrt{a^2-x^2}-\frac{x^2}{\sqrt{a^2-x^2}}+\frac{a^2}{2}\frac{a}{\sqrt{a^2-x^2}}\frac{1}{a}$

=$\frac{1}{2}\frac{a^2-x^2-x^2}{\sqrt{a^2-x^2}}+\frac{a^2}{2\sqrt{a^2-x^2}}$

=$\frac{a^2-2x^2}{2\sqrt{a^2-x^2}}+\frac{a^2}{2\sqrt{a^2-x^2}}$

=$\frac{a^2-2x^2+a^2}{2\sqrt{a^2-x^2}}$

=$\sqrt{a^2-x^2}{\sqrt{a^2-x^2}}$

=$\sqrt{a^2-x^2}$

Answer 2

In the attachment I have answered this problem.

I have applied chain rule to find derivative of the given function.

See the attachment for detailed solution.