Question

let (f(x)=7x^{2/5}-4x^{3/7}). evaluate each of

1. (f(3)=)
2. (f(5)=)

Explanation:

Step1: Apply power - rule for differentiation

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

$$\begin{align*} f^\prime(x)&=\frac{14}{5}x^{-\frac{3}{5}}-\frac{12}{7}x^{-\frac{4}{7}}\\ &=\frac{14}{5x^{3/5}}-\frac{12}{7x^{4/7}} \end{align*}$$

\]

Step2: Evaluate $f^\prime(3)$

Substitute $x = 3$ into $f^\prime(x)$:
\[

$$\begin{align*} f^\prime(3)&=\frac{14}{5\times3^{3/5}}-\frac{12}{7\times3^{4/7}}\\ &=\frac{14}{5\times\sqrt[5]{27}}-\frac{12}{7\times\sqrt[7]{81}} \end{align*}$$

\]
\[

$$\begin{align*} \frac{14}{5\times\sqrt[5]{27}}&\approx\frac{14}{5\times2.008}\\ &=\frac{14}{10.04}\approx1.394\\ \frac{12}{7\times\sqrt[7]{81}}&\approx\frac{12}{7\times2.167}\\ &=\frac{12}{15.169}\approx0.791 \end{align*}$$

\]
$f^\prime(3)\approx1.394 - 0.791=0.603$

Step3: Evaluate $f^\prime(5)$

Substitute $x = 5$ into $f^\prime(x)$:
\[

$$\begin{align*} f^\prime(5)&=\frac{14}{5\times5^{3/5}}-\frac{12}{7\times5^{4/7}}\\ &=\frac{14}{5\times\sqrt[5]{125}}-\frac{12}{7\times\sqrt[7]{625}} \end{align*}$$

\]
\[

$$\begin{align*} \frac{14}{5\times\sqrt[5]{125}}&=\frac{14}{5\times2.6265}\approx1.066\\ \frac{12}{7\times\sqrt[7]{625}}&=\frac{12}{7\times2.776}\approx0.617 \end{align*}$$

\]
$f^\prime(5)\approx1.066 - 0.617 = 0.449$

Answer:

1. $f^\prime(3)\approx0.603$
2. $f^\prime(5)\approx0.449$