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

-1/sqrt(1-9x)

Explanation:

This is the answer

Answer 2

• The derivative of the function is:

`f'(x)= (-1)/(2√(9-x))`

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

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

Explanation:

The function f(x) is given by:

`f(x)=√(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
`√(x)` is defined when x≥0.

Hence,
`√(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)  (f(x+h)-f(x))/(h)`

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

`f'(x)= \lim_(h \to 0) (√(9-(x+h))-√(9-x))/(h)\\\\i.e.\\\\f'(x)= \lim_(h \to 0) (√(9-(x+h))-√(9-x))/(h)* (√(9-(x+h))+√(9-x))/(√(9-(x+h))+√(9-x))\\\\\\f'(x)= \lim_(h \to 0) ((√(9-(x+h))-√(9-x))(√(9-(x+h))+√(9-x)))/((√(9-(x+h))+√(9-x))* h)\\\\\\f'(x)= \lim_(h \to 0) (9-(x+h)-(9-x))/((√(9-(x+h))+√(9-x))* h)`

Since,

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

Hence, we have:

`f'(x)= \lim_(h \to 0) (-h)/((√(9-(x+h))+√(9-x))* h)\\\\\\f'(x)= \lim_(h \to 0) (-1)/((√(9-(x+h))+√(9-x)))\\\\\\i.e.\\\\\\f'(x)= (-1)/(2√(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`