I⨍ ⨍ (x)=ax+a2ax, then ⨍ '(a)=……

Solution:

$G i v e n f (x) = \sqrt{a x} + \frac{a^{2}}{\sqrt{a x}} d i f f e r e n t i a t e w . r . t o x o n b o t h s i d e s ⨍^{'} (x)= \sqrt{a} × \frac{1}{2 \sqrt{x}} +a \sqrt{a} (- \frac{1}{2}) x^{- 3 / 2} s u b x = a o n b o t h s i d e s {⇒⨍}^{'} (a)= \sqrt{a} × \frac{1}{2 \sqrt{a}} - \frac{1}{2} a \sqrt{a} a^{- 3 / 2} = \frac{1}{2} - \frac{1}{2} =0.$

$I ⨍ ⨍ (x)= \sqrt{ax} + \frac{a^{2}}{\sqrt{ax},}$ then $⨍^{'} (a)=……$

Solution:

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