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
Answers
Answered by
13
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}\} \frac{d}{dx}\{\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}sin^{-1}\frac{x}{a}\}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5C%7B%5Cfrac%7Bx%7D%7B2%7D%5Csqrt%7Ba%5E2-x%5E2%7D%2B%5Cfrac%7Ba%5E2%7D%7B2%7Dsin%5E%7B-1%7D%5Cfrac%7Bx%7D%7Ba%7D%5C%7D)
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{a^2}{2}sin^{-1}\frac{x}{a}\}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5C%7B%5Cfrac%7Bx%7D%7B2%7D%5Csqrt%7Ba%5E2-x%5E2%7D%2B%5Cfrac%7Ba%5E2%7D%7B2%7Dsin%5E%7B-1%7D%5Cfrac%7Bx%7D%7Ba%7D%5C%7D)
=![\frac{d}{dx}\{\frac{x}{2}\sqrt{a^2-x^2}\}+\frac{d}{dx}\{\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}\}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5C%7B%5Cfrac%7Bx%7D%7B2%7D%5Csqrt%7Ba%5E2-x%5E2%7D%5C%7D%2B%5Cfrac%7Bd%7D%7Bdx%7D%5C%7B%5Cfrac%7Ba%5E2%7D%7B2%7Dsin%5E%7B-1%7D%5Cfrac%7Bx%7D%7Ba%7D%5C%7D)
=![\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{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}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%5Csqrt%7Ba%5E2-x%5E2%7D%5Cfrac%7Bdx%7D%7Bdx%7D%2Bx%5Cfrac%7Bd%5Csqrt%7Ba%5E2-x%5E2%7D%7D%7Bdx%7D%5D%2B%5Cfrac%7Ba%5E2%7D%7B2%7D%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D%7D%7D%5Cfrac%7Bd%5C%7B%5Cfrac%7Bx%7D%7Ba%7D%5C%7D%7D%7Bdx%7D)
=![\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}[\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}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B%5Csqrt%7Ba%5E2-x%5E2%7D-%5Cfrac%7Bx%5E2%7D%7B%5Csqrt%7Ba%5E2-x%5E2%7D%7D%2B%5Cfrac%7Ba%5E2%7D%7B2%7D%5Cfrac%7Ba%7D%7B%5Csqrt%7Ba%5E2-x%5E2%7D%7D%5Cfrac%7B1%7D%7Ba%7D)
=![\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{1}{2}\frac{a^2-x^2-x^2}{\sqrt{a^2-x^2}}+\frac{a^2}{2\sqrt{a^2-x^2}}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Ba%5E2-x%5E2-x%5E2%7D%7B%5Csqrt%7Ba%5E2-x%5E2%7D%7D%2B%5Cfrac%7Ba%5E2%7D%7B2%5Csqrt%7Ba%5E2-x%5E2%7D%7D)
=![\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}{2\sqrt{a^2-x^2}}+\frac{a^2}{2\sqrt{a^2-x^2}}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E2-2x%5E2%7D%7B2%5Csqrt%7Ba%5E2-x%5E2%7D%7D%2B%5Cfrac%7Ba%5E2%7D%7B2%5Csqrt%7Ba%5E2-x%5E2%7D%7D)
=![\frac{a^2-2x^2+a^2}{2\sqrt{a^2-x^2}} \frac{a^2-2x^2+a^2}{2\sqrt{a^2-x^2}}](https://tex.z-dn.net/?f=%5Cfrac%7Ba%5E2-2x%5E2%2Ba%5E2%7D%7B2%5Csqrt%7Ba%5E2-x%5E2%7D%7D)
=![\sqrt{a^2-x^2}{\sqrt{a^2-x^2}} \sqrt{a^2-x^2}{\sqrt{a^2-x^2}}](https://tex.z-dn.net/?f=%5Csqrt%7Ba%5E2-x%5E2%7D%7B%5Csqrt%7Ba%5E2-x%5E2%7D%7D)
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rohitkumargupta:
nice potential bhaiaya ji:-)
Answered by
1
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.
Attachments:
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