Quotient rule derivative and evaluation

Consider the function $g(x) = \dfrac{2x - 1}{x + 3}$ .

Find $g'(x)$ using the quotient rule, then evaluate $g'(0)$ .

Hint

Write $g(x) = \dfrac{f(x)}{h(x)}$ with $f(x) = 2x-1$ and $h(x) = x+3$ . Apply the quotient rule $g'(x) = \dfrac{f'(x)h(x) - f(x)h'(x)}{[h(x)]^2}$ , being careful to keep the order of subtraction correct in the numerator.

Step-by-step solution

Let $f(x) = 2x - 1$ and $h(x) = x + 3$ , so $f'(x) = 2$ and $h'(x) = 1$ . By the quotient rule, $g'(x) = \dfrac{f'(x)h(x) - f(x)h'(x)}{[h(x)]^2} = \dfrac{2(x+3) - (2x-1)(1)}{(x+3)^2}$ .

Expanding the numerator: $2(x+3) - (2x-1) = 2x + 6 - 2x + 1 = 7$ . So $g'(x) = \dfrac{7}{(x+3)^2}$ . Evaluating at $x=0$ gives $g'(0) = \dfrac{7}{9}$ .

Quotient rule derivative and evaluation

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