Find the derivative of each function using the limit definition. $\newline$ a. $\newline$ $f(x)=x^{2}+3x-5$

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Q. Find the derivative of each function using the limit definition.

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f(x)=x^{2}+3x-5

Write Limit Definition: Write down the limit definition of the derivative. $\newline$ The derivative of a function $f(x)$ at a point $x$ is given by the limit as $h$ approaches $0$ of the difference quotient: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ .
Apply to Function: Apply the limit definition to the function $f(x) = x^2 + 3x - 5$ . We need to calculate the limit as $h$ approaches $0$ of the expression: $(f(x+h) - f(x))/h = ((x+h)^2 + 3(x+h) - 5 - (x^2 + 3x - 5))/h$ .
Expand Numerator: Expand the numerator of the difference quotient. $\newline$ Expand $(x+h)^2$ and $3(x+h)$ in the numerator to get: $(x^2 + 2xh + h^2 + 3x + 3h - 5 - x^2 - 3x + 5)/h$ .
Simplify Numerator: Simplify the numerator by canceling out like terms. The terms $x^2$ , $3x$ , and $-5$ cancel out with $-x^2$ , $-3x$ , and $+5$ , respectively, leaving us with: $(2xh + h^2 + 3h)/h$ .
Factor Out $h$ : Factor out $h$ from the numerator. $\newline$ Factor $h$ from each term in the numerator to get: $\frac{h(2x + h + 3)}{h}$ .
Cancel $h$ : Simplify the difference quotient by canceling $h$ . Cancel $h$ in the numerator and denominator to get: $2x + h + 3$ .
Take Limit: Take the limit as $h$ approaches $0$ . Now we need to find the limit of $2x + h + 3$ as $h$ approaches $0$ , which is simply $2x + 3$ since the term involving $h$ vanishes.
Write Final Derivative: Write down the final derivative. $\newline$ The derivative of the function $f(x) = x^2 + 3x - 5$ is $f'(x) = 2x + 3$ .