Question

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

Explanation:

Step1: Apply power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y'=anx^{n - 1}$. For $f(x)=5x^{\frac{1}{5}}-4x^{\frac{5}{7}}$, we have $f'(x)=5\times\frac{1}{5}x^{\frac{1}{5}-1}-4\times\frac{5}{7}x^{\frac{5}{7}-1}$.
$f'(x)=x^{-\frac{4}{5}}-\frac{20}{7}x^{-\frac{2}{7}}$

Step2: Evaluate $f'(1)$

Substitute $x = 1$ into $f'(x)$. Since $1$ to any non - zero power is $1$, we have $f'(1)=1^{-\frac{4}{5}}-\frac{20}{7}\times1^{-\frac{2}{7}}$.
$f'(1)=1-\frac{20}{7}=-\frac{13}{7}$

Step3: Evaluate $f'(4)$

Substitute $x = 4$ into $f'(x)$. First, $4^{-\frac{4}{5}}=\frac{1}{4^{\frac{4}{5}}}=\frac{1}{(4^{\frac{1}{5}})^4}$ and $4^{-\frac{2}{7}}=\frac{1}{4^{\frac{2}{7}}}=\frac{1}{(4^{\frac{1}{7}})^2}$.
$f'(4)=4^{-\frac{4}{5}}-\frac{20}{7}\times4^{-\frac{2}{7}}=\frac{1}{4^{\frac{4}{5}}}-\frac{20}{7\times4^{\frac{2}{7}}}$
$4^{\frac{4}{5}} = 4^{\frac{28}{35}}$ and $4^{\frac{2}{7}}=4^{\frac{10}{35}}$.
$4^{\frac{4}{5}}=\sqrt[5]{4^4}=\sqrt[5]{256}$ and $4^{\frac{2}{7}}=\sqrt[7]{16}$.
$f'(4)=\frac{1}{\sqrt[5]{256}}-\frac{20}{7\sqrt[7]{16}}$
$1. f'(1)=-\frac{13}{7}$
$2. f'(4)=\frac{1}{4^{\frac{4}{5}}}-\frac{20}{7\times4^{\frac{2}{7}}}$

Answer:

1. $-\frac{13}{7}$
2. $\frac{1}{4^{\frac{4}{5}}}-\frac{20}{7\times4^{\frac{2}{7}}}$