Differentiating Inverse Trigonometric Functions - AP Calculus BC

  1. What is the chain rule application for y=sec⁡−1(g(x))y = \sec^{-1}(g(x))?
    g′(x)∣g(x)∣(g(x))2−1\frac{g'(x)}{|g(x)|\sqrt{(g(x))^2-1}}. General chain rule pattern for inverse secant composition.
  2. Find ddx(sin⁡−1(2x))\frac{d}{dx}(\sin^{-1}(2x)).
    21−(2x)2\frac{2}{\sqrt{1-(2x)^2}}. Chain rule with ddx(2x)=2\frac{d}{dx}(2x) = 2 applied to sin⁡−1\sin^{-1} formula.
  3. Find ddx(cos⁡−1(4x3))\frac{d}{dx}(\cos^{-1}(4x^3)).
    −12x21−(4x3)2-\frac{12x^2}{\sqrt{1-(4x^3)^2}}. Chain rule with ddx(4x3)=12x2\frac{d}{dx}(4x^3) = 12x^2 applied to cos⁡−1\cos^{-1}.
  4. Find ddx(tan⁡−1(3x))\frac{d}{dx}(\tan^{-1}(\sqrt{3x})).
    323x(1+3x)\frac{3}{2\sqrt{3x}(1+3x)}. Chain rule with ddx(3x)=323x\frac{d}{dx}(\sqrt{3x}) = \frac{3}{2\sqrt{3x}} applied to tan⁡−1\tan^{-1}.
  5. Find ddx(cot⁡−1(2x))\frac{d}{dx}(\cot^{-1}(\sqrt{2x})).
    −122x(1+2x)-\frac{1}{2\sqrt{2x}(1+2x)}. Chain rule with ddx(2x)=12x\frac{d}{dx}(\sqrt{2x}) = \frac{1}{\sqrt{2x}} applied to cot⁡−1\cot^{-1}.
  6. State the derivative of sin⁡−1(x)\sin^{-1}(x).
    11−x2\frac{1}{\sqrt{1-x^2}}. Basic derivative formula for inverse sine function.
  7. Evaluate ddx(cot⁡−1(x))\frac{d}{dx}(\cot^{-1}(x)) at x=1x = 1.
    −12-\frac{1}{2}. Substitute x=1x = 1 into −11+x2-\frac{1}{1+x^2}.
  8. State the derivative of cos⁡−1(x)\cos^{-1}(x).
    −11−x2-\frac{1}{\sqrt{1-x^2}}. Basic derivative formula for inverse cosine function.
  9. Identify the derivative of y=cot⁡−1(x)y = \cot^{-1}(\sqrt{x}).
    −12x(1+x)-\frac{1}{2\sqrt{x}(1+x)}. Chain rule with ddx(x)=12x\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} applied to cot⁡−1\cot^{-1}.
  10. Find ddx(csc⁡−1(2x3))\frac{d}{dx}(\csc^{-1}(2x^3)).
    −6x2∣2x3∣(2x3)2−1-\frac{6x^2}{|2x^3|\sqrt{(2x^3)^2-1}}. Chain rule with ddx(2x3)=6x2\frac{d}{dx}(2x^3) = 6x^2 applied to csc⁡−1\csc^{-1}.
  11. What is the chain rule application for y=cos⁡−1(g(x))y = \cos^{-1}(g(x))?
    −g′(x)1−(g(x))2-\frac{g'(x)}{\sqrt{1-(g(x))^2}}. General chain rule pattern for inverse cosine composition.
  12. Identify the derivative of y=tan⁡−1(x)y = \tan^{-1}(\sqrt{x}).
    12x(1+x)\frac{1}{2\sqrt{x}(1+x)}. Chain rule with ddx(x)=12x\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} applied to tan⁡−1\tan^{-1}.
  13. Evaluate ddx(csc⁡−1(x))\frac{d}{dx}(\csc^{-1}(x)) at x=2x = 2.
    −123-\frac{1}{2\sqrt{3}}. Substitute x=2x = 2 into −1∣x∣x2−1-\frac{1}{|x|\sqrt{x^2-1}}.
  14. Identify the derivative of y=sec⁡−1(x4)y = \sec^{-1}(x^4).
    4x3∣x4∣x8−1\frac{4x^3}{|x^4|\sqrt{x^8-1}}. Chain rule with ddx(x4)=4x3\frac{d}{dx}(x^4) = 4x^3 applied to sec⁡−1\sec^{-1} formula.
  15. Identify the derivative of y=cos⁡−1(x3)y = \cos^{-1}(x^3).
    −3x21−x6-\frac{3x^2}{\sqrt{1-x^6}}. Chain rule with ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2 applied to cos⁡−1\cos^{-1} formula.
  16. Find ddx(xcos⁡−1(x))\frac{d}{dx}(x \cos^{-1}(x)).
    cos⁡−1(x)−x1−x2\cos^{-1}(x) - \frac{x}{\sqrt{1-x^2}}. Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv' applied to xcos⁡−1(x)x\cos^{-1}(x).
  17. Find ddx(cos⁡−1(3x))\frac{d}{dx}(\cos^{-1}(3x)).
    −31−(3x)2-\frac{3}{\sqrt{1-(3x)^2}}. Chain rule with ddx(3x)=3\frac{d}{dx}(3x) = 3 applied to cos⁡−1\cos^{-1} formula.
  18. Identify the derivative of y=sin⁡−1(x2)y = \sin^{-1}(x^2).
    2x1−x4\frac{2x}{\sqrt{1-x^4}}. Chain rule with ddx(x2)=2x\frac{d}{dx}(x^2) = 2x applied to sin⁡−1\sin^{-1} formula.
  19. Evaluate ddx(tan⁡−1(x))\frac{d}{dx}(\tan^{-1}(x)) at x=1x = 1.
    12\frac{1}{2}. Substitute x=1x = 1 into 11+x2\frac{1}{1+x^2}.
  20. Evaluate ddx(sin⁡−1(x))\frac{d}{dx}(\sin^{-1}(x)) at x=12x = \frac{1}{2}.
    233\frac{2\sqrt{3}}{3}. Substitute x=12x = \frac{1}{2} into 11−x2\frac{1}{\sqrt{1-x^2}}.
  21. What is the chain rule application for y=csc⁡−1(g(x))y = \csc^{-1}(g(x))?
    −g′(x)∣g(x)∣(g(x))2−1-\frac{g'(x)}{|g(x)|\sqrt{(g(x))^2-1}}. General chain rule pattern for inverse cosecant composition.
  22. Evaluate ddx(cos⁡−1(x))\frac{d}{dx}(\cos^{-1}(x)) at x=12x = \frac{1}{2}.
    −233-\frac{2\sqrt{3}}{3}. Substitute x=12x = \frac{1}{2} into −11−x2-\frac{1}{\sqrt{1-x^2}}.
  23. Find ddx(sec⁡−1(6x))\frac{d}{dx}(\sec^{-1}(6x)).
    6∣6x∣(6x)2−1\frac{6}{|6x|\sqrt{(6x)^2-1}}. Chain rule with ddx(6x)=6\frac{d}{dx}(6x) = 6 applied to sec⁡−1\sec^{-1} formula.
  24. Find ddx(xsin⁡−1(x))\frac{d}{dx}(x \sin^{-1}(x)).
    sin⁡−1(x)+x1−x2\sin^{-1}(x) + \frac{x}{\sqrt{1-x^2}}. Product rule: (uv)′=u′v+uv′(uv)' = u'v + uv' applied to xsin⁡−1(x)x\sin^{-1}(x).
  25. State the derivative of tan⁡−1(x)\tan^{-1}(x).
    11+x2\frac{1}{1+x^2}. Basic derivative formula for inverse tangent function.
  26. Find ddx(sec⁡−1(5x2))\frac{d}{dx}(\sec^{-1}(5x^2)).
    10x∣5x2∣(5x2)2−1\frac{10x}{|5x^2|\sqrt{(5x^2)^2-1}}. Chain rule with ddx(5x2)=10x\frac{d}{dx}(5x^2) = 10x applied to sec⁡−1\sec^{-1}.
  27. What is the chain rule application for y=tan⁡−1(g(x))y = \tan^{-1}(g(x))?
    g′(x)1+(g(x))2\frac{g'(x)}{1+(g(x))^2}. General chain rule pattern for inverse tangent composition.
  28. Find ddx(sin⁡−1(3x2))\frac{d}{dx}(\sin^{-1}(3x^2)).
    6x1−(3x2)2\frac{6x}{\sqrt{1-(3x^2)^2}}. Chain rule with ddx(3x2)=6x\frac{d}{dx}(3x^2) = 6x applied to sin⁡−1\sin^{-1}.
  29. Find ddx(tan⁡−1(5x))\frac{d}{dx}(\tan^{-1}(5x)).
    51+(5x)2\frac{5}{1+(5x)^2}. Chain rule with ddx(5x)=5\frac{d}{dx}(5x) = 5 applied to tan⁡−1\tan^{-1} formula.
  30. What is the chain rule application for y=cot⁡−1(g(x))y = \cot^{-1}(g(x))?
    −g′(x)1+(g(x))2-\frac{g'(x)}{1+(g(x))^2}. General chain rule pattern for inverse cotangent composition.
  31. Find ddx(cot⁡−1(4x))\frac{d}{dx}(\cot^{-1}(4x)).
    −41+(4x)2-\frac{4}{1+(4x)^2}. Chain rule with ddx(4x)=4\frac{d}{dx}(4x) = 4 applied to cot⁡−1\cot^{-1} formula.
  32. Identify the derivative of y=csc⁡−1(x5)y = \csc^{-1}(x^5).
    −5x4∣x5∣x10−1-\frac{5x^4}{|x^5|\sqrt{x^{10}-1}}. Chain rule with ddx(x5)=5x4\frac{d}{dx}(x^5) = 5x^4 applied to csc⁡−1\csc^{-1} formula.
  33. State the derivative of cot⁡−1(x)\cot^{-1}(x).
    −11+x2-\frac{1}{1+x^2}. Basic derivative formula for inverse cotangent function.
  34. What is the chain rule application for y=sin⁡−1(g(x))y = \sin^{-1}(g(x))?
    g′(x)1−(g(x))2\frac{g'(x)}{\sqrt{1-(g(x))^2}}. General chain rule pattern for inverse sine composition.
  35. State the derivative of sec⁡−1(x)\sec^{-1}(x).
    1∣x∣x2−1\frac{1}{|x|\sqrt{x^2-1}}. Basic derivative formula for inverse secant function.
  36. Find ddx(csc⁡−1(7x))\frac{d}{dx}(\csc^{-1}(7x)).
    −7∣7x∣(7x)2−1-\frac{7}{|7x|\sqrt{(7x)^2-1}}. Chain rule with ddx(7x)=7\frac{d}{dx}(7x) = 7 applied to csc⁡−1\csc^{-1} formula.
  37. Evaluate ddx(sec⁡−1(x))\frac{d}{dx}(\sec^{-1}(x)) at x=2x = 2.
    123\frac{1}{2\sqrt{3}}. Substitute x=2x = 2 into 1∣x∣x2−1\frac{1}{|x|\sqrt{x^2-1}}.
  38. State the derivative of csc⁡−1(x)\csc^{-1}(x).
    −1∣x∣x2−1-\frac{1}{|x|\sqrt{x^2-1}}. Basic derivative formula for inverse cosecant function.

Study these flashcards