Derivatives of Trigonometry and Logarithmic Functions - AP Calculus BC

  1. Find the derivative: y=ln xy = \text{ln } x.
    dydx=1x\frac{dy}{dx} = \frac{1}{x}. Apply the derivative rule for ln x\text{ln } x.
  2. Find the second derivative of y=ln xy = \text{ln } x.
    d2ydx2=−1x2\frac{d^2y}{dx^2} = -\frac{1}{x^2}. Differentiate f′(x)=1xf'(x) = \frac{1}{x} to get f′′(x)=−1x2f''(x) = -\frac{1}{x^2}.
  3. Find the derivative of f(x)=ln (5x)f(x) = \text{ln }(5x).
    f′(x)=1xf'(x) = \frac{1}{x}. Constant multiple rule: ddx[ln(5x)]=15x⋅5=1x\frac{d}{dx}[\text{ln}(5x)] = \frac{1}{5x} \cdot 5 = \frac{1}{x}.
  4. Find the second derivative of y=exy = e^x.
    d2ydx2=ex\frac{d^2y}{dx^2} = e^x. Differentiate f′(x)=exf'(x) = e^x to get f′′(x)=exf''(x) = e^x.
  5. Determine the derivative of f(x)=xexf(x) = xe^x.
    f′(x)=ex+xexf'(x) = e^x + xe^x. Use product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'.
  6. Determine f′(x)f'(x) if f(x)=cos x+sin xf(x) = \text{cos } x + \text{sin } x.
    f′(x)=−sin x+cos xf'(x) = -\text{sin } x + \text{cos } x. Use sum rule: derivative of each term separately.
  7. Differentiate f(x)=sin xf(x) = \text{sin } x at x=π2x = \frac{\text{π}}{2}.
    f′(π2)=0f'\big(\frac{\text{π}}{2}\big) = 0. f′(x)=cos xf'(x) = \text{cos } x, so f′(π2)=cos(π2)=0f'\big(\frac{\text{π}}{2}\big) = \text{cos}\big(\frac{\text{π}}{2}\big) = 0.
  8. What is the derivative of sin x\text{sin } x?
    cos x\text{cos } x. Basic derivative rule for sine function.
  9. Find the derivative of f(x)=e2xf(x) = e^{2x}.
    f′(x)=2e2xf'(x) = 2e^{2x}. Use chain rule with inner function 2x2x.
  10. Find the derivative of f(x)=cos (3x)f(x) = \text{cos }(3x).
    f′(x)=−3sin (3x)f'(x) = -3\text{sin }(3x). Use chain rule with inner function 3x3x.
  11. Find the derivative: y=cos xy = \text{cos } x.
    dydx=−sin x\frac{dy}{dx} = -\text{sin } x. Apply the derivative rule for cos x\text{cos } x.
  12. Find the derivative of f(x)=sin (2x)f(x) = \text{sin }(2x).
    f′(x)=2cos (2x)f'(x) = 2\text{cos }(2x). Use chain rule with inner function 2x2x.
  13. What is the derivative of exe^x?
    exe^x. The exponential function is its own derivative.
  14. Determine the derivative of f(x)=xcos xf(x) = x\text{cos } x.
    f′(x)=cos x−xsin xf'(x) = \text{cos } x - x\text{sin } x. Use product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'.
  15. Find the derivative: y=exy = e^x.
    dydx=ex\frac{dy}{dx} = e^x. Apply the derivative rule for exe^x.
  16. Determine f′(x)f'(x) if f(x)=cos⁡x+sin⁡xf(x) = \cos x + \sin x.
    f′(x)=−sin⁡x+cos⁡xf'(x) = -\sin x + \cos x. Use sum rule: derivative of each term separately.
  17. Evaluate the derivative: f(x)=ln xf(x) = \text{ln } x at x=2x = 2.
    f′(2)=12f'(2) = \frac{1}{2}. f′(x)=1xf'(x) = \frac{1}{x}, so f′(2)=12f'(2) = \frac{1}{2}.
  18. Determine f′(x)f'(x) if f(x)=ex+ln xf(x) = e^x + \text{ln } x.
    f′(x)=ex+1xf'(x) = e^x + \frac{1}{x}. Use sum rule: derivative of each term separately.
  19. Evaluate the derivative: f(x)=exf(x) = e^x at x=0x = 0.
    f′(0)=1f'(0) = 1. f′(x)=exf'(x) = e^x, so f′(0)=e0=1f'(0) = e^0 = 1.
  20. What is the derivative of ln x\text{ln } x?
    1x\frac{1}{x}. Standard derivative of natural logarithm.
  21. Determine the derivative of f(x)=xln xf(x) = x\text{ln } x.
    f′(x)=ln x+1f'(x) = \text{ln } x + 1. Use product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'.
  22. Determine the derivative of f(x)=xsin xf(x) = x\text{sin } x.
    f′(x)=sin x+xcos xf'(x) = \text{sin } x + x\text{cos } x. Use product rule: (uv)′=u′v+uv′(uv)' = u'v + uv'.
  23. Differentiate f(x)=3cos x−2sin xf(x) = 3\text{cos } x - 2\text{sin } x.
    f′(x)=−3sin x−2cos xf'(x) = -3\text{sin } x - 2\text{cos } x. Use sum/difference rule with constant multiples.
  24. Evaluate f′(x)f'(x) for f(x)=ex−ln xf(x) = e^x - \text{ln } x at x=1x = 1.
    f′(1)=e−1f'(1) = e - 1. f′(x)=ex−1xf'(x) = e^x - \frac{1}{x}, so f′(1)=e−1f'(1) = e - 1.
  25. Evaluate the derivative: f(x)=exf(x) = e^x at x=ln 2x = \text{ln } 2.
    f′(ln 2)=2f'(\text{ln } 2) = 2. f′(x)=exf'(x) = e^x, so f′(ln 2)=eln 2=2f'(\text{ln } 2) = e^{\text{ln } 2} = 2.
  26. Find the slope of the tangent to y=ln xy = \text{ln } x at x=4x = 4.
    Slope = 14\frac{1}{4}. The slope equals the derivative at the point.
  27. What is the derivative of cos x\text{cos } x?
    −sin x-\text{sin } x. Basic derivative rule for cosine function.
  28. Differentiate f(x)=cos xf(x) = \text{cos } x at x=0x = 0.
    f′(0)=0f'(0) = 0. f′(x)=−sin xf'(x) = -\text{sin } x, so f′(0)=−sin (0)=0f'(0) = -\text{sin }(0) = 0.
  29. Differentiate f(x)=5ex+7ln xf(x) = 5e^x + 7\text{ln } x.
    f′(x)=5ex+7xf'(x) = 5e^x + \frac{7}{x}. Use sum rule with constant multiples.
  30. Find the derivative of f(x)=sin⁡(2x)f(x) = \sin(2x).
    f′(x)=2cos⁡(2x)f'(x) = 2\cos(2x). Use chain rule with inner function 2x2x.
  31. Evaluate the derivative: f(x)=sin xf(x) = \text{sin } x at x=0x = 0.
    f′(0)=1f'(0) = 1. f′(x)=cos xf'(x) = \text{cos } x, so f′(0)=cos(0)=1f'(0) = \text{cos}(0) = 1.
  32. Evaluate f′(x)f'(x) for f(x)=cos x−exf(x) = \text{cos } x - e^x at x=0x = 0.
    f′(0)=−1f'(0) = -1. f′(x)=−sin x−exf'(x) = -\text{sin } x - e^x, so f′(0)=0−1=−1f'(0) = 0 - 1 = -1.
  33. Find the second derivative of y=cos xy = \text{cos } x.
    d2ydx2=−cos x\frac{d^2y}{dx^2} = -\text{cos } x. Differentiate f′(x)=−sin xf'(x) = -\text{sin } x to get f′′(x)=−cos xf''(x) = -\text{cos } x.
  34. Find the second derivative of y=sin xy = \text{sin } x.
    d2ydx2=−sin x\frac{d^2y}{dx^2} = -\text{sin } x. Differentiate f′(x)=cos xf'(x) = \text{cos } x to get f′′(x)=−sin xf''(x) = -\text{sin } x.
  35. Evaluate the derivative: f(x)=ln xf(x) = \text{ln } x at x=1x = 1.
    f′(1)=1f'(1) = 1. f′(x)=1xf'(x) = \frac{1}{x}, so f′(1)=11=1f'(1) = \frac{1}{1} = 1.
  36. Find the derivative: y=sin xy = \text{sin } x.
    dydx=cos x\frac{dy}{dx} = \text{cos } x. Apply the derivative rule for sin x\text{sin } x.
  37. Differentiate f(x)=exf(x) = e^x at x=1x = 1.
    f′(1)=ef'(1) = e. f′(x)=exf'(x) = e^x, so f′(1)=e1=ef'(1) = e^1 = e.
  38. Differentiate f(x)=ln xf(x) = \text{ln } x at x=ex = e.
    f′(e)=1ef'(e) = \frac{1}{e}. f′(x)=1xf'(x) = \frac{1}{x}, so f′(e)=1ef'(e) = \frac{1}{e}.
  39. Find the slope of the tangent to y=exy = e^x at x=2x = 2.
    Slope = e2e^2. The slope equals the derivative at the point.
  40. Evaluate the derivative: f(x)=cos xf(x) = \text{cos } x at x=π2x = \frac{\text{π}}{2}.
    f′(π2)=−1f'\big(\frac{\text{π}}{2}\big) = -1. f′(x)=−sin xf'(x) = -\text{sin } x, so f′(π2)=−sin(π2)=−1f'\big(\frac{\text{π}}{2}\big) = -\text{sin}\big(\frac{\text{π}}{2}\big) = -1.

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