Question

question 2
2 pts
if $f(x)=\frac{4x^{3}-2x^{2}+7x - 1}{sqrt{x}}$, then $f(x)=$
$12x^{2}-4x + 7-\frac{1}{2}x^{-1/2}$
$\frac{12x^{2}-4x + 6}{2sqrt{x}}$
$10x^{3/2}-3x^{1/2}+\frac{7}{2}x^{-1/2}+\frac{1}{2}x^{-3/2}$
$\frac{12x^{2}-4x + 7}{\frac{1}{2}x^{-1/2}}$

question 3
2 pts
if $f(x)=\frac{4x^{3}-2x^{2}+7x - 1}{e^{x}}$, then $f(x)=$
$-(12x^{2}-4x + 7)e^{-x}$
$\frac{12x^{2}-4x + 7}{e^{x}}$
$\frac{(4x^{3}-2x^{2}+7x - 1)e^{x}-e^{x}(12x^{2}-4x + 7)}{(e^{x})^{2}}$
$\frac{e^{x}(12x^{2}-4x + 7)-(4x^{3}-2x^{2}+7x - 1)e^{x}}{(e^{x})^{2}}$

Explanation:

Question 2

Step1: Rewrite the function

Rewrite $f(x)=\frac{4x^{3}-2x^{2}+7x - 1}{\sqrt{x}}=4x^{\frac{5}{2}}-2x^{\frac{3}{2}}+7x^{\frac{1}{2}}-x^{-\frac{1}{2}}$.

Step2: Apply power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $y^\prime=nx^{n - 1}$.
For $y = 4x^{\frac{5}{2}}$, $y^\prime=4\times\frac{5}{2}x^{\frac{5}{2}-1}=10x^{\frac{3}{2}}$.
For $y=-2x^{\frac{3}{2}}$, $y^\prime=-2\times\frac{3}{2}x^{\frac{3}{2}-1}=-3x^{\frac{1}{2}}$.
For $y = 7x^{\frac{1}{2}}$, $y^\prime=7\times\frac{1}{2}x^{\frac{1}{2}-1}=\frac{7}{2}x^{-\frac{1}{2}}$.
For $y=-x^{-\frac{1}{2}}$, $y^\prime=-(-\frac{1}{2})x^{-\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{3}{2}}$.
So $f^\prime(x)=10x^{\frac{3}{2}}-3x^{\frac{1}{2}}+\frac{7}{2}x^{-\frac{1}{2}}+\frac{1}{2}x^{-\frac{3}{2}}$.

Step1: Identify the quotient rule

The quotient rule for a function $y=\frac{u}{v}$ is $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = 4x^{3}-2x^{2}+7x - 1$, $u^\prime=12x^{2}-4x + 7$, and $v = e^{x}$, $v^\prime=e^{x}$.

Step2: Apply the quotient rule

Substitute $u$, $u^\prime$, $v$, and $v^\prime$ into the quotient - rule formula.
$f^\prime(x)=\frac{(12x^{2}-4x + 7)e^{x}-(4x^{3}-2x^{2}+7x - 1)e^{x}}{(e^{x})^{2}}=\frac{(4x^{3}-2x^{2}+7x - 1)e^{x}-e^{x}(12x^{2}-4x + 7)}{(e^{x})^{2}}$.

Answer:

$10x^{\frac{3}{2}}-3x^{\frac{1}{2}}+\frac{7}{2}x^{-\frac{1}{2}}+\frac{1}{2}x^{-\frac{3}{2}}$

Question 3