Calculating Higher-Order Derivatives - AP Calculus BC

1. What is the fourth derivative of $f(x) = x^4$?

   $f^{(4)}(x) = 24$. Fourth derivative of $x^4$ gives the factorial $4! = 24$.

2. What is the second derivative of $f(x) = \text{sin}^2(x)$?

   $f''(x) = 2\text{cos}(2x)$. Use double angle identity: $\sin^2(x) = \frac{1-\cos(2x)}{2}$.

3. What is the second derivative of $f(x) = e^x$?

   $f''(x) = e^x$. Derivative of $e^x$ is always $e^x$.

4. Compute the third derivative of $f(x) = \text{e}^{-2x}$.

   $f'''(x) = -8\text{e}^{-2x}$. Chain rule with $e^{-2x}$: coefficient becomes $(-2)^3 = -8$.

5. What is the second derivative of $f(x) = x^2 \text{e}^x$?

   $f''(x) = (x^2 + 4x + 2)\text{e}^x$. Product rule applied twice to $x^2 e^x$.

6. State the second derivative of $f(x) = \text{cos}(x)$.

   $f''(x) = -\text{cos}(x)$. Second derivative of $\cos(x)$ follows trig cycle pattern.

7. Find the third derivative of $f(x) = \frac{1}{3}x^3$.

   $f'''(x) = 2$. Third derivative eliminates the coefficient $\frac{1}{3}$ leaving constant $2$.

8. What is the second derivative of $f(x) = \text{e}^{3x}$?

   $f''(x) = 9\text{e}^{3x}$. Chain rule with $e^{3x}$: coefficient becomes $3^2 = 9$.

9. Find the second derivative of $f(x) = \text{e}^{-x}$.

   $f''(x) = \text{e}^{-x}$. Derivative of $e^{-x}$ multiplies by $(-1)^2 = 1$.

10. What is the third derivative of $f(x) = \text{sin}(2x)$?

    $f'''(x) = -8\text{sin}(2x)$. Chain rule with $\sin(2x)$: coefficient becomes $(-2)^3 = -8$.

11. What is the second derivative of $f(x) = \text{e}^{2x}$?

    $f''(x) = 4\text{e}^{2x}$. Chain rule with exponential: derivative multiplies by the inner function coefficient squared.

12. Compute the fourth derivative of $f(x) = \text{cos}(3x)$.

    $f^{(4)}(x) = 81\text{cos}(3x)$. Fourth derivative of $\cos(3x)$ involves $3^4 = 81$ and returns to cosine.

13. Find the third derivative of $f(x) = 5x^4 + 3x^3 - x$.

    $f'''(x) = 120x + 18$. Apply power rule to each term and differentiate three times.

14. What is the second derivative of $f(x) = x^3$?

    $f''(x) = 6x$. Apply power rule twice: $f'(x) = 3x^2$, then $f''(x) = 6x$.

15. What is the second derivative of $f(x) = x \text{ln}(x)$?

    $f''(x) = \frac{1}{x}$. Use product rule: $f'(x) = \ln(x) + 1$, then $f''(x) = \frac{1}{x}$.

16. State the second derivative of $f(x) = \text{tan}(x)$.

    $f''(x) = 2\text{sec}^2(x)\text{tan}(x)$. First derivative is $\sec^2(x)$, apply chain rule again.

17. Find the second derivative of $f(x) = e^{-x}$.

    $f''(x) = e^{-x}$. Derivative of $e^{-x}$ multiplies by $(-1)^2 = 1$.

18. What is the second derivative of $f(x) = \text{ln}(x^2)$?

    $f''(x) = -\frac{2}{x^2}$. Use chain rule: $\ln(x^2) = 2\ln(x)$, so derivative doubles.

19. Find the fourth derivative of $f(x) = x^4 - 3x^2 + 2$.

    $f^{(4)}(x) = 24$. Fourth derivative of $x^4$ term is $24$, other terms become zero.

20. Compute the third derivative of $f(x) = \text{sin}(x)$.

    $f'''(x) = -\text{cos}(x)$. Trig derivatives cycle: $\sin \to \cos \to -\sin \to -\cos$.

21. What is the second derivative of $f(x) = \frac{1}{x}$?

    $f''(x) = \frac{2}{x^3}$. Rewrite as $x^{-1}$, then apply power rule twice.

22. Compute the third derivative of $f(x) = \frac{1}{x^2}$.

    $f'''(x) = \frac{6}{x^5}$. Rewrite as $x^{-2}$ and apply power rule three times.

23. Compute the third derivative of $f(x) = x^2 \text{sin}(x)$.

    $f'''(x) = -2(3\text{cos}(x) + x\text{sin}(x))$. Use product rule repeatedly on $x^2 \sin(x)$.

24. What is the fourth derivative of $f(x) = x^4 - x^2 + 1$?

    $f^{(4)}(x) = 24$. Fourth derivative eliminates lower-order terms, leaving only $x^4$ coefficient.

25. State the formula for the second derivative of $f(x) = \text{ln}(x)$.

    $f''(x) = -\frac{1}{x^2}$. Standard formula for logarithmic second derivative.

26. What is the second derivative of $f(x) = \text{arcsin}(x)$?

    $f''(x) = \frac{x}{(1-x^2)^{3/2}}$. Inverse trig derivatives involve radical expressions in denominators.

27. What is the second derivative of $f(x) = \text{ln}(x)$?

    $f''(x) = -\frac{1}{x^2}$. First derivative is $\frac{1}{x}$, second is $-\frac{1}{x^2}$.

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