Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. f(t) = sqrt 9-x

Answer:The derivative of the function is:        [tex]f'(x)= \dfrac{-1}{2\sqrt{9-x}}[/tex]The domain of the function is:  [tex]x\leq 9[/tex]and the domain of the derivative function is: [tex]x\leq 9[/tex]Step-by-step explanation:The function f(x) is given by:    [tex]f(x)=\sqrt{9-x}[/tex]The domain of the function is the possible values of x where the function is defined.We know that the square root function [tex]\sqrt{x}[/tex] is defined when x≥0.Hence, [tex]\sqrt{9-x}[/tex] will be defined when [tex]9-x\geq 0\\\\i.e.\\\\x\leq 9[/tex]Hence, the domain of the function f(x) is: [tex]x\leq 9[/tex]Also, the definition of derivative of x is given by:[tex]f'(x)= \lim_{h \to 0}  \dfrac{f(x+h)-f(x)}{h}[/tex]Hence, here by putting the value of the function we get:[tex]f'(x)= \lim_{h \to 0} \dfrac{\sqrt{9-(x+h)}-\sqrt{9-x}}{h}\\\\i.e.\\\\f'(x)= \lim_{h \to 0} \dfrac{\sqrt{9-(x+h)}-\sqrt{9-x}}{h}\times \dfrac{\sqrt{9-(x+h)}+\sqrt{9-x}}{\sqrt{9-(x+h)}+\sqrt{9-x}}\\\\\\f'(x)= \lim_{h \to 0} \dfrac{(\sqrt{9-(x+h)}-\sqrt{9-x})(\sqrt{9-(x+h)}+\sqrt{9-x})}{(\sqrt{9-(x+h)}+\sqrt{9-x})\times h}\\\\\\f'(x)= \lim_{h \to 0} \dfrac{9-(x+h)-(9-x)}{(\sqrt{9-(x+h)}+\sqrt{9-x})\times h}[/tex]Since,[tex](a-b)(a+b)=a^2-b^2[/tex]Hence, we have:[tex]f'(x)= \lim_{h \to 0} \dfrac{-h}{(\sqrt{9-(x+h)}+\sqrt{9-x})\times h}\\\\\\f'(x)= \lim_{h \to 0} \dfrac{-1}{(\sqrt{9-(x+h)}+\sqrt{9-x})}\\\\\\i.e.\\\\\\f'(x)= \dfrac{-1}{2\sqrt{9-x}}[/tex]Since, the domain of the derivative function is equal to the derivative of the square root function.Also, the domain of the square root function is: [tex]x\leq 9[/tex]Hence, domain of the derivative function is:  [tex]x\leq 9[/tex]