Question

hw 10 - product and quotient rules section 2.5: problem 4 (1 point)
let ( f(x)=(7x - 7x^{3})(5+sqrt{x})). find ( f(x)).
( f(x)=square)

Explanation:

Step1: Recall product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Let $u(x)=7x - 7x^{3}$ and $v(x)=5+\sqrt{x}=5 + x^{\frac{1}{2}}$.

Step2: Find $u'(x)$

Differentiate $u(x)=7x - 7x^{3}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $u'(x)=\frac{d}{dx}(7x)-\frac{d}{dx}(7x^{3})=7-21x^{2}$.

Step3: Find $v'(x)$

Differentiate $v(x)=5 + x^{\frac{1}{2}}$ with respect to $x$. $\frac{d}{dx}(5)=0$ and $\frac{d}{dx}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}$. So $v'(x)=\frac{1}{2\sqrt{x}}$.

Step4: Apply product - rule

$f'(x)=u'(x)v(x)+u(x)v'(x)=(7 - 21x^{2})(5+\sqrt{x})+(7x - 7x^{3})\frac{1}{2\sqrt{x}}$.
Expand the first term: $(7 - 21x^{2})(5+\sqrt{x})=35 + 7\sqrt{x}-105x^{2}-21x^{\frac{5}{2}}$.
Expand the second term: $(7x - 7x^{3})\frac{1}{2\sqrt{x}}=\frac{7x}{2\sqrt{x}}-\frac{7x^{3}}{2\sqrt{x}}=\frac{7}{2}\sqrt{x}-\frac{7}{2}x^{\frac{5}{2}}$.
Combine like - terms:
\[

$$\begin{align*} f'(x)&=35 + 7\sqrt{x}-105x^{2}-21x^{\frac{5}{2}}+\frac{7}{2}\sqrt{x}-\frac{7}{2}x^{\frac{5}{2}}\\ &=35+(7 + \frac{7}{2})\sqrt{x}-105x^{2}-(21+\frac{7}{2})x^{\frac{5}{2}}\\ &=35+\frac{14 + 7}{2}\sqrt{x}-105x^{2}-\frac{42 + 7}{2}x^{\frac{5}{2}}\\ &=35+\frac{21}{2}\sqrt{x}-105x^{2}-\frac{49}{2}x^{\frac{5}{2}} \end{align*}$$

\]

Answer:

$35+\frac{21}{2}\sqrt{x}-105x^{2}-\frac{49}{2}x^{\frac{5}{2}}$