Question

derivatives of basic functions: problem 2 (1 point) results for this submission 2 of the answers are not correct. suppose that $f(x)=1x^{3/5}-2x^{2/7}$. evaluate each of the following: $f(2)=$ $f(4)=$ note: you can earn partial credit on this problem. preview my answers submit answers your score was recorded. your score was successfully sent to canvas. you have attempted this problem 2 times. you received a score of 0% for this attempt. your overall recorded score is 0%. you have unlimited attempts remaining.

Explanation:

Step1: Apply power - rule for derivatives

The power - rule states that if $y = ax^n$, then $y'=anx^{n - 1}$. For $f(x)=x^{3/5}-2x^{2/7}$, the derivative $f'(x)$ is $f'(x)=\frac{3}{5}x^{\frac{3}{5}-1}-2\times\frac{2}{7}x^{\frac{2}{7}-1}=\frac{3}{5}x^{-\frac{2}{5}}-\frac{4}{7}x^{-\frac{5}{7}}$.

Step2: Evaluate $f'(2)$

Substitute $x = 2$ into $f'(x)$:
\[

$$\begin{align*} f'(2)&=\frac{3}{5}\times2^{-\frac{2}{5}}-\frac{4}{7}\times2^{-\frac{5}{7}}\\ &=\frac{3}{5\times2^{\frac{2}{5}}}-\frac{4}{7\times2^{\frac{5}{7}}} \end{align*}$$

\]
\[

$$\begin{align*} 2^{\frac{2}{5}}&=\sqrt[5]{4}\approx1.32\\ 2^{\frac{5}{7}}&=2^{5/7}\approx1.75 \end{align*}$$

\]
\[

$$\begin{align*} f'(2)&=\frac{3}{5\times1.32}-\frac{4}{7\times1.75}\\ &=\frac{3}{6.6}-\frac{4}{12.25}\\ &=\frac{30}{66}-\frac{4}{12.25}\\ &=\frac{5}{11}-\frac{4}{12.25}\\ &=\frac{5\times12.25 - 4\times11}{11\times12.25}\\ &=\frac{61.25-44}{134.75}\\ &=\frac{17.25}{134.75}\approx0.128 \end{align*}$$

\]

Step3: Evaluate $f'(4)$

Substitute $x = 4$ into $f'(x)$:
\[

$$\begin{align*} f'(4)&=\frac{3}{5}\times4^{-\frac{2}{5}}-\frac{4}{7}\times4^{-\frac{5}{7}}\\ &=\frac{3}{5\times4^{\frac{2}{5}}}-\frac{4}{7\times4^{\frac{5}{7}}} \end{align*}$$

\]
\[

$$\begin{align*} 4^{\frac{2}{5}}&=\sqrt[5]{16}\approx1.74\\ 4^{\frac{5}{7}}&=4^{5/7}\approx2.64 \end{align*}$$

\]
\[

$$\begin{align*} f'(4)&=\frac{3}{5\times1.74}-\frac{4}{7\times2.64}\\ &=\frac{3}{8.7}-\frac{4}{18.48}\\ &=\frac{30}{87}-\frac{4}{18.48}\\ &=\frac{10}{29}-\frac{4}{18.48}\\ &=\frac{10\times18.48-4\times29}{29\times18.48}\\ &=\frac{184.8 - 116}{535.92}\\ &=\frac{68.8}{535.92}\approx0.129 \end{align*}$$

\]

Answer:

$f'(2)\approx0.128$, $f'(4)\approx0.129$