Question

given: $f(x) = \frac{2x}{5x^2 + 8}$. find the slope of the tangent line to this curve at $x = 1$.\
$\circ \frac{16}{169}$\
$\circ \frac{6}{169}$\
$\circ \frac{46}{169}$\
$\circ \frac{21}{169}$\
$\circ -\frac{8}{169}$\
$\circ$ none of the above.\
question number 5. (4.00 points)

Explanation:

Step1: Recall the Quotient Rule

The quotient rule for differentiation states that if \( f(x) = \frac{g(x)}{h(x)} \), then \( f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \). Here, \( g(x) = 2x \) and \( h(x) = 5x^2 + 8 \).

Step2: Find \( g'(x) \) and \( h'(x) \)

• For \( g(x) = 2x \), the derivative \( g'(x) = 2 \) (using the power rule \( \frac{d}{dx}(x^n) = nx^{n - 1} \), here \( n = 1 \)).
• For \( h(x) = 5x^2 + 8 \), the derivative \( h'(x) = 10x \) (using the power rule: \( \frac{d}{dx}(5x^2)=10x \) and \( \frac{d}{dx}(8) = 0 \)).

Step3: Apply the Quotient Rule

Substitute \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \) into the quotient rule formula:
\[
f'(x) = \frac{(2)(5x^2 + 8) - (2x)(10x)}{(5x^2 + 8)^2}
\]
Simplify the numerator:
\[

$$\begin{align*} (2)(5x^2 + 8) - (2x)(10x) &= 10x^2 + 16 - 20x^2\\ &= -10x^2 + 16 \end{align*}$$

\]
So, \( f'(x) = \frac{-10x^2 + 16}{(5x^2 + 8)^2} \).

Step4: Evaluate \( f'(x) \) at \( x = 1 \)

Substitute \( x = 1 \) into \( f'(x) \):

• Numerator: \( -10(1)^2 + 16 = -10 + 16 = 6 \)
• Denominator: \( (5(1)^2 + 8)^2 = (5 + 8)^2 = 13^2 = 169 \)

Thus, \( f'(1) = \frac{6}{169} \).

Answer:

\(\frac{6}{169}\) (corresponding to the option " \(\boldsymbol{\frac{6}{169}}\)")