MCQ
જો $f(x) = \sqrt {ax} + {{{a^2}} \over {\sqrt {ax} }},$ તો $f'(a) = $
- A$-1$
- B$1$
- ✓$0$
- D$a$
==> $f'(x) = \frac{{\sqrt a }}{{2\sqrt x }} + \frac{{{a^2}}}{{\sqrt a }}\left( {\frac{{ - 1}}{2}{x^{ - 3/2}}} \right)$
==> $f'(x) = \frac{{\sqrt a }}{{2\sqrt x }} - \frac{{{a^2}}}{{2\sqrt a }}{x^{ - 3/2}}$
==> $f'(a) = \frac{{\sqrt a }}{{2\sqrt a }} - \frac{{{a^2}}}{{2\sqrt a \,.\,{a^{3/2}}}}$
==>$f'(a) = \frac{1}{2} - \frac{{{a^2}}}{{2{a^2}}} = 0$.
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