Question

differentiate the given function.

$y = x^2 \sqrt{4x - 1}$

$y = \square$

(type an exact answer, using radicals as needed.)

Explanation:

Step1: Identify the product rule

We have \( y = x^2 \sqrt{4x - 1} \), which is a product of two functions \( u = x^2 \) and \( v=\sqrt{4x - 1}=(4x - 1)^{\frac{1}{2}} \). The product rule states that \( y'=u'v + uv' \).

Step2: Differentiate \( u = x^2 \)

Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), we get \( u' = 2x \).

Step3: Differentiate \( v=(4x - 1)^{\frac{1}{2}} \)

Using the chain rule \( \frac{d}{dx}(f(g(x)))=f'(g(x))\cdot g'(x) \). Let \( f(u)=u^{\frac{1}{2}} \) and \( g(x)=4x - 1 \). Then \( f'(u)=\frac{1}{2}u^{-\frac{1}{2}} \) and \( g'(x) = 4 \). So \( v'=\frac{1}{2}(4x - 1)^{-\frac{1}{2}}\cdot4=\frac{2}{\sqrt{4x - 1}} \).

Step4: Apply the product rule

\( y'=u'v+uv'=2x\cdot\sqrt{4x - 1}+x^2\cdot\frac{2}{\sqrt{4x - 1}} \)

Step5: Combine the terms

To combine the terms, we get a common denominator of \( \sqrt{4x - 1} \):
\[

$$\begin{align*} y'&=\frac{2x(4x - 1)+2x^2}{\sqrt{4x - 1}}\\ &=\frac{8x^2-2x + 2x^2}{\sqrt{4x - 1}}\\ &=\frac{10x^2-2x}{\sqrt{4x - 1}}\\ &=\frac{2x(5x - 1)}{\sqrt{4x - 1}} \end{align*}$$

\]

Answer:

\(\dfrac{2x(5x - 1)}{\sqrt{4x - 1}}\)