Question
21. DETAILS MY NOTES LARCALC12 2.2.052.MI. Find the derivative of the function. \[ \begin{array}{l} f(x)=\frac{2}{\sqrt[3]{x}}+9 \cos (x) \\ f^{\prime}(x)=\square \end{array} \]
Ask by Wells Marshall. in the United States
Mar 03,2025
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Answer
The derivative of the function \( f(x) = \frac{2}{\sqrt[3]{x}} + 9 \cos(x) \) is:
\[
f'(x) = -\frac{2}{3}x^{-\frac{4}{3}} - 9 \sin(x)
\]
Solution
We are given the function
f(x) = 2/∛x + 9 cos(x)
Step 1. Rewrite ∛x as x^(1/3) so that 2/∛x becomes 2x^(–1/3).
Thus, f(x) = 2x^(–1/3) + 9 cos(x).
Step 2. Differentiate term by term.
• For 2x^(–1/3):
Use the power rule d/dx[x^n] = n x^(n–1).
d/dx[2x^(–1/3)] = 2 * (–1/3)x^(–1/3–1) = –(2/3)x^(–4/3).
• For 9 cos(x):
We know d/dx[cos(x)] = –sin(x).
Thus, d/dx[9 cos(x)] = 9 * (–sin(x)) = –9 sin(x).
Step 3. Combine the results:
f '(x) = –(2/3)x^(–4/3) – 9 sin(x).
Thus, the derivative is:
f '(x) = –(2/3)x^(–4/3) – 9 sin(x).
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Simplify this solution Bonus Knowledge
To find the derivative of the function \( f(x) = \frac{2}{\sqrt[3]{x}} + 9 \cos(x) \), we can proceed by using the power rule and the chain rule. We first rewrite \( \frac{2}{\sqrt[3]{x}} \) as \( 2x^{-\frac{1}{3}} \). Now differentiating, we have: \[ f'(x) = 2 \cdot -\frac{1}{3} x^{-\frac{4}{3}} + 9 \cdot (-\sin(x)) = -\frac{2}{3}x^{-\frac{4}{3}} - 9\sin(x). \] Therefore, the final derivative is: \[ f^{\prime}(x) = -\frac{2}{3\sqrt[3]{x^4}} - 9\sin(x). \] Now, make sure to apply proper rules and stay vigilant with signs when differentiating trigonometric functions, as it's a common pitfall for many learners! Happy differentiating!
