Question

Consider the function.SX-9(0, 9/2)(a) Find the value of the derivative of the function at the given point.(b) Choose which differentiation rule(s) you used to find the derivative. (Select all that apply)Quotient rulepower ruleproduct rule

Answer 1

Solution:

Given:

$\begin{gathered} g(x)=\frac{5x-9}{x^2-2} \\ \\ Let\text{ }g(x)=\frac{u}{v} \\ u=5x-9 \\ v=x^2-2 \end{gathered}$

Applying the quotient rule,

$\begin{gathered} g^{\prime}(x)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2} \\ v=x^2-2 \\ \frac{du}{dx}=5 \\ u=5x-9 \\ \frac{dv}{dx}=2x \\ \\ Hence, \\ g^{\prime}(x)=\frac{(x^2-2)(5)-(5x-9)(2x)}{(x^2-2)^2} \end{gathered}$

Expanding and simplifying further;

$\begin{gathered} g^{\prime}(x)=\frac{5x^2-10-(10x^2-18x)}{(x^2-2)^2} \\ g^{\prime}(x)=\frac{5x^2-10x^2-10+18x}{(x^2-2)^2} \\ g^{\prime}(x)=\frac{-5x^2+18x-10}{(x^2-2)^2} \end{gathered}$

Part A:

g'(0)

$\begin{gathered} g^{\prime}(0)\text{ is the value of g\lparen x\rparen when x = 0} \\ Substitute\text{ x = 0 into g'\lparen x\rparen} \\ g^{\prime}(0)=\frac{-5(0^2)+18(0)-10}{(0^2-2)^2} \\ g^{\prime}(0)=\frac{-10}{(-2)^2} \\ g^{\prime}(0)=\frac{-10}{4} \\ g^{\prime}(0)=-2.5 \end{gathered}$

Therefore, g'(0) = -2.5

Part B:

The differentiation rules used to find the derivative are the quotient rule and power rule.