Question

attempt 1: 10 attempts remaining. find the derivative of $f(x) = \sqrt{x}(3x^3 + x^2 - 2x - 5)$. $f(x) = $ submit answer next item

Explanation:

Step1: Rewrite the function

First, rewrite \( f(x)=\sqrt{x}(3x^{3}+x^{2}-2x - 5) \) as \( f(x)=x^{\frac{1}{2}}(3x^{3}+x^{2}-2x - 5) \).

Step2: Apply the product rule

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

• Find \( u^{\prime}(x) \): Using the power rule \( \frac{d}{dx}(x^{n})=nx^{n - 1} \), we have \( u^{\prime}(x)=\frac{1}{2}x^{-\frac{1}{2}} \).
• Find \( v^{\prime}(x) \): Using the power rule for each term, \( v^{\prime}(x)=9x^{2}+2x - 2 \).

Step3: Substitute into the product rule

\( f^{\prime}(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)=\frac{1}{2}x^{-\frac{1}{2}}(3x^{3}+x^{2}-2x - 5)+x^{\frac{1}{2}}(9x^{2}+2x - 2) \).

Step4: Simplify the expression

First, rewrite \( x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}} \) and \( x^{\frac{1}{2}}=\sqrt{x} \):
\( f^{\prime}(x)=\frac{3x^{3}+x^{2}-2x - 5}{2\sqrt{x}}+\sqrt{x}(9x^{2}+2x - 2) \)
To combine the terms, get a common denominator of \( 2\sqrt{x} \):
\( \sqrt{x}(9x^{2}+2x - 2)=\frac{2x(9x^{2}+2x - 2)}{2\sqrt{x}}=\frac{18x^{3}+4x^{2}-4x}{2\sqrt{x}} \)
Now add the two fractions:
\( f^{\prime}(x)=\frac{3x^{3}+x^{2}-2x - 5+18x^{3}+4x^{2}-4x}{2\sqrt{x}}=\frac{21x^{3}+5x^{2}-6x - 5}{2\sqrt{x}} \)
We can also rewrite this as \( \frac{21x^{3}+5x^{2}-6x - 5}{2\sqrt{x}}=\frac{21}{2}x^{\frac{5}{2}}+\frac{5}{2}x^{\frac{3}{2}}-3x^{\frac{1}{2}}-\frac{5}{2}x^{-\frac{1}{2}} \) (by dividing each term in the numerator by \( 2\sqrt{x}=2x^{\frac{1}{2}} \) and using the rule \( \frac{x^{m}}{x^{n}}=x^{m - n} \)).

Answer:

\( \frac{21x^{3}+5x^{2}-6x - 5}{2\sqrt{x}} \) (or \( \frac{21}{2}x^{\frac{5}{2}}+\frac{5}{2}x^{\frac{3}{2}}-3\sqrt{x}-\frac{5}{2\sqrt{x}} \))