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

Question

Accepted Answer

# 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. $f^\prime(5)\approx0.449$

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Question

Explanation:

Step1: Apply power - rule for differentiation

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

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

Answer: