Question

5 multiple choice: do one - a - day

1. if (f(x)=xsqrt{2x - 3}), then (f(x)=

(a) \frac{3x - 3}{sqrt{2x - 3}}
(b) \frac{x}{sqrt{2x - 3}}
(c) \frac{1}{sqrt{2x - 3}}
(d) \frac{-x + 3}{sqrt{2x - 3}}
(e) \frac{5x - 6}{2sqrt{2x - 3}}
no calcu

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = x$ 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 product - rule

$f'(x)=u'v+uv'$. Substitute $u = x$, $u' = 1$, $v=\sqrt{2x - 3}$, and $v'=\frac{1}{\sqrt{2x - 3}}$ into the product - rule formula: $f'(x)=1\cdot\sqrt{2x - 3}+x\cdot\frac{1}{\sqrt{2x - 3}}=\frac{2x - 3+x}{\sqrt{2x - 3}}=\frac{3x - 3}{\sqrt{2x - 3}}$.

Answer:

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