Question

find the derivative of the function using the limit process.

$f(x)=x^{2}+x - 9$

$f(x)=lim_{delta x
ightarrow0}\frac{f(x+delta x)-f(x)}{delta x}$

$=lim_{delta x
ightarrow0}\frac{square-(x^{2}+x - 9)}{delta x}$

$=square$

Explanation:

Step1: Compute \( f(x+\Delta x) \)

\( f(x+\Delta x) = (x+\Delta x)^2 + (x+\Delta x) - 9 = x^2 + 2x\Delta x + (\Delta x)^2 + x + \Delta x - 9 \)

Step2: Calculate \( f(x+\Delta x)-f(x) \)

\( [x^2 + 2x\Delta x + (\Delta x)^2 + x + \Delta x - 9] - (x^2 + x - 9) = 2x\Delta x + (\Delta x)^2 + \Delta x \)

Step3: Divide by \( \Delta x \)

\( \frac{2x\Delta x + (\Delta x)^2 + \Delta x}{\Delta x} = 2x + \Delta x + 1 \)

Step4: Take limit as \( \Delta x \to 0 \)

\( \lim_{\Delta x \to 0} (2x + \Delta x + 1) = 2x + 1 \)

Answer:

\( 2x + 1 \)