Answered

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Find f '(a).
f(x) = [tex]\frac{1}{\sqrt{x} +5}[/tex]

Sagot :

Medunno13

Answer:

[tex]f'(a)=\frac{1}{2a^{3/2}+20a+50\sqrt{a}}[/tex]

Step-by-step explanation:

[tex]f'(x)=\frac{\frac{d}{dx} (\sqrt{x}+5)(1)-(\sqrt{x}+5)\frac{d}{dx}(1)}{(\sqrt{x}+5)^2} \\ \\ =\frac{\frac{1}{2\sqrt{x}}}{(\sqrt{x}+5)^2} \\ \\ =\frac{\frac{1}{2\sqrt{x}}}{x+10\sqrt{x}+25} \\ \\ =\frac{1}{2x^{3/2}+20x+50\sqrt{x}} \\ \\ \therefore f'(a)=\frac{1}{2a^{3/2}+20a+50\sqrt{a}}[/tex]

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