Question

attempt 1: 10 attempts remaining. find the derivative of f(x) = √x (5x³ + x² - 2x - 7). f(x) = submit answer next item

Explanation:

Step1: Rewrite the function

First, rewrite \(\sqrt{x}\) as \(x^{\frac{1}{2}}\). So the function \(f(x)=x^{\frac{1}{2}}(5x^{3}+x^{2}-2x - 7)\).

Step2: Apply the product rule

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

• Find \(u'\): Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), for \(u=x^{\frac{1}{2}}\), \(u'=\frac{1}{2}x^{-\frac{1}{2}}\).
• Find \(v'\): Differentiate \(v = 5x^{3}+x^{2}-2x - 7\) term - by - term. \(\frac{d}{dx}(5x^{3})=15x^{2}\), \(\frac{d}{dx}(x^{2}) = 2x\), \(\frac{d}{dx}(-2x)=-2\), \(\frac{d}{dx}(-7) = 0\). So \(v'=15x^{2}+2x - 2\).

Step3: Substitute into the product rule

\(f'(x)=u'v+uv'\)
\(=\frac{1}{2}x^{-\frac{1}{2}}(5x^{3}+x^{2}-2x - 7)+x^{\frac{1}{2}}(15x^{2}+2x - 2)\)

Step4: Simplify the expression

First, distribute \(x^{\frac{1}{2}}\) in the second term: \(x^{\frac{1}{2}}(15x^{2}+2x - 2)=15x^{\frac{5}{2}}+2x^{\frac{3}{2}}-2x^{\frac{1}{2}}\)
Distribute \(\frac{1}{2}x^{-\frac{1}{2}}\) in the first term:
\(\frac{1}{2}x^{-\frac{1}{2}}(5x^{3})=\frac{5}{2}x^{\frac{5}{2}}\), \(\frac{1}{2}x^{-\frac{1}{2}}(x^{2})=\frac{1}{2}x^{\frac{3}{2}}\), \(\frac{1}{2}x^{-\frac{1}{2}}(-2x)=-x^{\frac{1}{2}}\), \(\frac{1}{2}x^{-\frac{1}{2}}(-7)=-\frac{7}{2}x^{-\frac{1}{2}}\)
Now, combine like terms:

• For \(x^{\frac{5}{2}}\) terms: \(\frac{5}{2}x^{\frac{5}{2}}+15x^{\frac{5}{2}}=\frac{5 + 30}{2}x^{\frac{5}{2}}=\frac{35}{2}x^{\frac{5}{2}}\)
• For \(x^{\frac{3}{2}}\) terms: \(\frac{1}{2}x^{\frac{3}{2}}+2x^{\frac{3}{2}}=\frac{1 + 4}{2}x^{\frac{3}{2}}=\frac{5}{2}x^{\frac{3}{2}}\)
• For \(x^{\frac{1}{2}}\) terms: \(-x^{\frac{1}{2}}-2x^{\frac{1}{2}}=-3x^{\frac{1}{2}}\)
• The remaining term: \(-\frac{7}{2}x^{-\frac{1}{2}}\)

So \(f'(x)=\frac{35}{2}x^{\frac{5}{2}}+\frac{5}{2}x^{\frac{3}{2}}-3x^{\frac{1}{2}}-\frac{7}{2}x^{-\frac{1}{2}}\)
We can also rewrite it with radicals:
\(f'(x)=\frac{35}{2}x^{2}\sqrt{x}+\frac{5}{2}x\sqrt{x}-3\sqrt{x}-\frac{7}{2\sqrt{x}}\)
Or combine over a common denominator:
Multiply each term by \(\frac{2\sqrt{x}}{2\sqrt{x}}\) to get a common denominator of \(2\sqrt{x}\):
\(f'(x)=\frac{35x^{3}+5x^{2}-6x - 7}{2\sqrt{x}}\) (by combining the terms: \(\frac{35}{2}x^{\frac{5}{2}}\times2\sqrt{x}=35x^{3}\), \(\frac{5}{2}x^{\frac{3}{2}}\times2\sqrt{x}=5x^{2}\), \(- 3x^{\frac{1}{2}}\times2\sqrt{x}=-6x\), \(-\frac{7}{2}x^{-\frac{1}{2}}\times2\sqrt{x}=-7\))

Answer:

\(f'(x)=\frac{35x^{3}+5x^{2}-6x - 7}{2\sqrt{x}}\) (or in the form with exponents \(\frac{35}{2}x^{\frac{5}{2}}+\frac{5}{2}x^{\frac{3}{2}}-3x^{\frac{1}{2}}-\frac{7}{2}x^{-\frac{1}{2}}\))