Question

attempt 1: 10 attempts remaining. let $f(x)=(5x - 8x^{3})(3+sqrt{x})$. find $f(x)$. $f(x)=$ submit answer next item answers your answer 01

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u(x)v(x)$, then $y^\prime=u^\prime(x)v(x)+u(x)v^\prime(x)$. Let $u(x)=5x - 8x^{3}$ and $v(x)=3+\sqrt{x}=3 + x^{\frac{1}{2}}$.

Step2: Find $u^\prime(x)$

Differentiate $u(x)$ term - by - term. Using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $u^\prime(x)=\frac{d}{dx}(5x)-\frac{d}{dx}(8x^{3})=5-24x^{2}$.

Step3: Find $v^\prime(x)$

Differentiate $v(x)$ term - by - term. $\frac{d}{dx}(3) = 0$ and $\frac{d}{dx}(x^{\frac{1}{2}})=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$. So $v^\prime(x)=\frac{1}{2\sqrt{x}}$.

Step4: Apply the product - rule formula

$f^\prime(x)=u^\prime(x)v(x)+u(x)v^\prime(x)=(5 - 24x^{2})(3+\sqrt{x})+(5x - 8x^{3})\frac{1}{2\sqrt{x}}$.
Expand the first part: $(5 - 24x^{2})(3+\sqrt{x})=15 + 5\sqrt{x}-72x^{2}-24x^{\frac{5}{2}}$.
Expand the second part: $(5x - 8x^{3})\frac{1}{2\sqrt{x}}=\frac{5x}{2\sqrt{x}}-\frac{8x^{3}}{2\sqrt{x}}=\frac{5}{2}\sqrt{x}-4x^{\frac{5}{2}}$.
Combine like terms:
\[

$$\begin{align*} f^\prime(x)&=15 + 5\sqrt{x}-72x^{2}-24x^{\frac{5}{2}}+\frac{5}{2}\sqrt{x}-4x^{\frac{5}{2}}\\ &=15+(5 + \frac{5}{2})\sqrt{x}-72x^{2}-(24 + 4)x^{\frac{5}{2}}\\ &=15+\frac{10 + 5}{2}\sqrt{x}-72x^{2}-28x^{\frac{5}{2}}\\ &=15+\frac{15}{2}\sqrt{x}-72x^{2}-28x^{\frac{5}{2}} \end{align*}$$

\]

Answer:

$15+\frac{15}{2}\sqrt{x}-72x^{2}-28x^{\frac{5}{2}}$