part ii: spiral material - keep the math fresh! 3. if f(x)=x√(2x - 3), then f(x)=(a) (3x - 3)/√(2x - 3) (b)…

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Here, $u = x$ and $v=\sqrt{2x - 3}=(2x - 3)^{\frac{1}{2}}$. First, find $u'$ and $v'$. The derivative of $u=x$ is $u' = 1$.

Step2: Find $v'$ using chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. For $v=(2x - 3)^{\frac{1}{2}}$, let $g(x)=2x - 3$ and $f(u)=u^{\frac{1}{2}}$. Then $f'(u)=\frac{1}{2}u^{-\frac{1}{2}}$ and $g'(x)=2$. So $v'=\frac{1}{2}(2x - 3)^{-\frac{1}{2}}\cdot2=(2x - 3)^{-\frac{1}{2}}=\frac{1}{\sqrt{2x - 3}}$.

Step3: Calculate $f'(x)$

Using the product - rule $f'(x)=u'v+uv'$, substitute $u = x$, $u' = 1$, $v=(2x - 3)^{\frac{1}{2}}$, and $v'=\frac{1}{\sqrt{2x - 3}}$: [ \begin{align*} f'(x)&=1\cdot\sqrt{2x - 3}+x\cdot\frac{1}{\sqrt{2x - 3}}\ &=\frac{2x - 3}{\sqrt{2x - 3}}+\frac{x}{\sqrt{2x - 3}}\ &=\frac{2x-3 + x}{\sqrt{2x - 3}}\ &=\frac{3x - 3}{\sqrt{2x - 3}} \end{align*} ]

Answer:

A. $\frac{3x - 3}{\sqrt{2x - 3}}$