To find the derivative of the given function $y = (3x^2 + 5x - 1) \sqrt{x}$, we can proceed by applying the product rule and then simplifying.

Given: $y = (3x^2 + 5x - 1) \sqrt{x}$

Rewrite $\sqrt{x}$ as $x^{1/2}$ for easier differentiation.

The function $y$ is a product of two functions:

1. $u(x) = 3x^2 + 5x - 1$
2. $v(x) = x^{1/2}$

The derivative of $y$ with respect to $x$ is: $y' = u'v + uv'$

Step 1: Differentiate $u(x) = 3x^2 + 5x - 1$

$u' = 6x + 5$

Step 2: Differentiate $v(x) = x^{1/2}$

Using the power rule, $v' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$.

Step 3: Substitute and Simplify

Now, substitute $u$, $u'$, $v$, and $v'$ into the product rule formula: $y' = (6x + 5)x^{1/2} + (3x^2 + 5x - 1) \frac{1}{2\sqrt{x}}$

Step 4: Simplify Each Term

1. The first term, $(6x + 5)x^{1/2}$, becomes: $(6x + 5)x^{1/2} = 6x^{3/2} + 5x^{1/2}$

2. The second term, $(3x^2 + 5x - 1) \frac{1}{2\sqrt{x}}$, becomes: $\frac{3x^2 + 5x - 1}{2\sqrt{x}} = \frac{3x^2}{2x^{1/2}} + \frac{5x}{2x^{1/2}} - \frac{1}{2x^{1/2}}$ Simplifying each component: $= \frac{3}{2}x^{3/2} + \frac{5}{2}x^{1/2} - \frac{1}{2}x^{-1/2}$

Step 5: Combine All Terms

Now combine all terms to get the final derivative: $y' = 6x^{3/2} + 5x^{1/2} + \frac{3}{2}x^{3/2} + \frac{5}{2}x^{1/2} - \frac{1}{2}x^{-1/2}$

Combine like terms: $y' = \left(6 + \frac{3}{2}\right)x^{3/2} + \left(5 + \frac{5}{2}\right)x^{1/2} - \frac{1}{2}x^{-1/2}$

Simplifying coefficients: $y' = \frac{15}{2}x^{3/2} + \frac{15}{2}x^{1/2} - \frac{1}{2}x^{-1/2}$

So, the derivative is: $y' = \frac{15}{2}x^{3/2} + \frac{15}{2}x^{1/2} - \frac{1}{2}x^{-1/2}$

Would you like further clarification on any part of this process?

Related Questions to Explore

1. How is the product rule applied to functions of different powers?
2. How can we simplify expressions involving fractional exponents?
3. What alternative methods could we use to find derivatives of products?
4. Why is it helpful to rewrite square roots as fractional exponents in calculus?
5. How does the chain rule apply if this were a composite function?

Tip

When dealing with square roots in derivatives, rewriting them as fractional exponents often simplifies the process of differentiation.