Question

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 1

Answer:

• The derivative of the function is:

       $f'(x)= \dfrac{-1}{2\sqrt{9-x}}$

• The domain of the function is:  $x\leq 9$

• and the domain of the derivative function is: $x\leq 9$

Step-by-step explanation:

The function f(x) is given by:

   $f(x)=\sqrt{9-x}$

The domain of the function is the possible values of x where the function is defined.

We know that the square root function $\sqrt{x}$ is defined when x≥0.

Hence, $\sqrt{9-x}$ will be defined when $9-x\geq 0\\\\i.e.\\\\x\leq 9$

Hence, the domain of the function f(x) is: $x\leq 9$

Also, the definition of derivative of x is given by:

$f'(x)= \lim_{h \to 0}  \dfrac{f(x+h)-f(x)}{h}$

Hence, here by putting the value of the function we get:

$f'(x)= \lim_{h \to 0} \dfrac{\sqrt{9-(x+h)}-\sqrt{9-x}}{h}\\\\i.e.\\\\f'(x)= \lim_{h \to 0} \dfrac{\sqrt{9-(x+h)}-\sqrt{9-x}}{h}\times \dfrac{\sqrt{9-(x+h)}+\sqrt{9-x}}{\sqrt{9-(x+h)}+\sqrt{9-x}}\\\\\\f'(x)= \lim_{h \to 0} \dfrac{(\sqrt{9-(x+h)}-\sqrt{9-x})(\sqrt{9-(x+h)}+\sqrt{9-x})}{(\sqrt{9-(x+h)}+\sqrt{9-x})\times h}\\\\\\f'(x)= \lim_{h \to 0} \dfrac{9-(x+h)-(9-x)}{(\sqrt{9-(x+h)}+\sqrt{9-x})\times h}$

Since,

$(a-b)(a+b)=a^2-b^2$

Hence, we have:

$f'(x)= \lim_{h \to 0} \dfrac{-h}{(\sqrt{9-(x+h)}+\sqrt{9-x})\times h}\\\\\\f'(x)= \lim_{h \to 0} \dfrac{-1}{(\sqrt{9-(x+h)}+\sqrt{9-x})}\\\\\\i.e.\\\\\\f'(x)= \dfrac{-1}{2\sqrt{9-x}}$

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: $x\leq 9$

Hence, domain of the derivative function is:  $x\leq 9$

Answer 2

Answer:

-1/sqrt(1-9x)

Step-by-step explanation:

This is the answer