Question

step 2
therefore, we have
y = \frac{7x^{2}+3x + 7}{sqrt{x}}=7x^{3/2}+3x^{1/2}+7x^{- 1/2}
now, since the function is a sum of power - functions, we can use the power rule to find its derivative. recall that according to the power rule, the derivative of (x^{n}), where (n) is any real number, is (nx^{n - 1}). to use the power rule, well have to calculate the following.
(\frac{3}{2}-1=\frac{1}{2})
(\frac{1}{2}-1 =-\frac{1}{2})
(-\frac{1}{2}-1=-\frac{3}{2})
applying the power rule we have
y=\frac{21}{2}x^{1/2}+\frac{3}{2}x^{-1/2}+left(-\frac{7}{2}
ight)x^{-3/2}
step 3
now we can conclude that
y=

Explanation:

Step1: Recall the power - rule derivative

The power - rule states that the derivative of $x^n$ is $nx^{n - 1}$. We have $y'=\frac{21}{2}x^{\frac{1}{2}}+\frac{3}{2}x^{-\frac{1}{2}}-\frac{7}{2}x^{-\frac{3}{2}}$.

Step2: Combine the terms

We can write it as a single fraction. First, find a common denominator, which is $2x^{\frac{3}{2}}$.
\[

$$\begin{align*} y'&=\frac{21x^{\frac{1}{2}}\cdot x^{\frac{3}{2}}+3x^{-\frac{1}{2}}\cdot x^{\frac{3}{2}}- 7x^{-\frac{3}{2}}\cdot x^{\frac{3}{2}}}{2x^{\frac{3}{2}}}\\ &=\frac{21x^{2}+3x - 7}{2x^{\frac{3}{2}}} \end{align*}$$

\]

Answer:

$\frac{21x^{2}+3x - 7}{2x^{\frac{3}{2}}}$