If f(x) = ax + a^2 ax , then f'(a) = - Vidyadip

MCQ

If $f(x) = \sqrt {ax} + {{{a^2}} \over {\sqrt {ax} }},$ then $f'(a) = $

A
$-1$
B
$1$
✓
$0$
D
$a$

✓

Answer

Correct option: C.

$0$

c
(c) $f(x) = \sqrt {ax} + \frac{{{a^2}}}{{\sqrt {ax} }},$ then

==> $f'(x) = \frac{{\sqrt a }}{{2\sqrt x }} + \frac{{{a^2}}}{{\sqrt a }}\left( {\frac{{ - 1}}{2}{x^{ - 3/2}}} \right)$

==> $f'(x) = \frac{{\sqrt a }}{{2\sqrt x }} - \frac{{{a^2}}}{{2\sqrt a }}{x^{ - 3/2}}$

==> $f'(a) = \frac{{\sqrt a }}{{2\sqrt a }} - \frac{{{a^2}}}{{2\sqrt a \,.\,{a^{3/2}}}}$

==>$f'(a) = \frac{1}{2} - \frac{{{a^2}}}{{2{a^2}}} = 0$.

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