hw 10 - product and quotient rules section 2.5: problem 9 (1 point) if (f(x)=\frac{7x + 3}{3x + 5}), find…

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $f(x)=\frac{u(x)}{v(x)}$, then $f^{\prime}(x)=\frac{u^{\prime}(x)v(x)-u(x)v^{\prime}(x)}{v^{2}(x)}$. Here, $u(x)=7x + 3$ and $v(x)=3x + 5$.

Step2: Find $u^{\prime}(x)$ and $v^{\prime}(x)$

Differentiate $u(x)=7x + 3$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax + b)=a$, we get $u^{\prime}(x)=7$. Differentiate $v(x)=3x + 5$ with respect to $x$, we get $v^{\prime}(x)=3$.

Step3: Apply the quotient - rule

Substitute $u(x)=7x + 3$, $u^{\prime}(x)=7$, $v(x)=3x + 5$, and $v^{\prime}(x)=3$ into the quotient - rule formula: [ \begin{align*} f^{\prime}(x)&=\frac{7(3x + 5)-(7x + 3)\times3}{(3x + 5)^{2}}\ &=\frac{21x+35-(21x + 9)}{(3x + 5)^{2}}\ &=\frac{21x+35 - 21x-9}{(3x + 5)^{2}}\ &=\frac{26}{(3x + 5)^{2}} \end{align*} ]

Answer:

$\frac{26}{(3x + 5)^{2}}$