More practice questions

5. Q5. Write log₂ 32, log₁₀ 0.001 and log₉ 27 as numbers.

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   2⁵ = 32, so log₂ 32 = 5. 10⁻³ = 0.001, so log₁₀ 0.001 = −3. 27 = 9^3/2, so log₉ 27 = 3/2 (or, by change of base, log₃ 27/log₃ 9 = 3/2).

6. Q6. Differentiate e^x log x, x > 0.

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   By the product rule: e^x log x + e^x/x = e^x(log x + 1/x).

7. Q7. Differentiate log (x + √(x² + 1)).

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   The derivative of the inside is 1 + x/√(x² + 1) = (√(x² + 1) + x)/√(x² + 1). Dividing by the inside leaves 1/√(x² + 1).

8. Q8. For which x is log (cos e^x) defined? Find its derivative there.

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   It needs cos e^x > 0: 0 < e^x < π/2, or 2nπ − π/2 < e^x < 2nπ + π/2 for n = 1, 2, 3, …. In terms of x: x < log (π/2), or log (2nπ − π/2) < x < log (2nπ + π/2). There the derivative is −e^x tan e^x.

9. Q9. Show that the tangent line to y = e^x at (0, 1) is y = x + 1, and that the tangent to y = log x at (1, 0) is its mirror image y = x − 1.

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   d/dx (e^x) = 1 at x = 0, so the tangent is y − 1 = 1(x − 0). d/dx (log x) = 1 at x = 1, so the tangent is y = x − 1. Swapping x and y in y = x + 1 gives x = y + 1, that is y = x − 1: the reflection in y = x.