Topic Review — Differentiation of Exponential and Logarithmic Functions — Solutions

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This review covers all lessons in this topic: differentiating e^x, composite exponential functions, ln(x), and composite log functions with applications. Questions are mixed in difficulty.

Review Questions

Find d/dx[e^5x].
5e^5x
Find d/dx[ln(4x)].
ln(4x) = ln 4 + ln x, so d/dx = 1/x
Differentiate y = e^x².
dy/dx = 2x ċ e^x²
Differentiate y = ln(2x + 5).
dy/dx = 2/(2x + 5)
Find d/dx[3e^−2x].
−6e^−2x
Find the gradient of y = ln(x² + 3) at x = 1.
dy/dx = 2x/(x²+3); at x=1: gradient = 2/4 = 1/2
Use log laws to differentiate y = ln(x³(x+2)²).
y = 3 ln x + 2 ln(x+2)
dy/dx = 3/x + 2/(x+2)
Find the equation of the tangent to y = e^2x at x = 0.
At x=0: y=1; dy/dx=2e^2x; at x=0: gradient=2
Tangent: y − 1 = 2(x − 0) ⇒ y = 2x + 1
Find the stationary point of y = e^−x + x and determine its nature.
dy/dx = −e^−x + 1 = 0 ⇒ e^−x = 1 ⇒ x = 0
y = 1 + 0 = 1; d²y/dx² = e^−x; at x=0: value = 1 > 0
Local minimum at (0, 1)
State the domain of y = ln(3 − x) and find dy/dx.
Domain: 3 − x > 0 ⇒ x < 3
dy/dx = −1/(3 − x)
Differentiate y = ln√(x+1) and simplify.
y = ½ ln(x+1)
dy/dx = 1/(2(x+1))
A quantity grows as Q = Q₀e^0.05t. Find dQ/dt and interpret it.
dQ/dt = 0.05Q₀e^0.05t = 0.05Q
The rate of growth is 5% of the current quantity — this is exponential growth.
Find the stationary points of f(x) = x² − 2 ln x and classify each.
Domain: x > 0
f′(x) = 2x − 2/x = 0 ⇒ 2x² = 2 ⇒ x = 1
f(1) = 1 − 0 = 1; f′′(x) = 2 + 2/x²; f′′(1) = 4 > 0
Local minimum at (1, 1)
Find d/dx[e^x − e^−x].
d/dx[e^x] = e^x; d/dx[−e^−x] = e^−x
Answer: e^x + e^−x
A particle’s position is given by x(t) = ln(t² + 1). Find the velocity at t = 2 and determine when the particle is stationary.
v(t) = x′(t) = 2t/(t²+1)
At t = 2: v = 4/5 = 0.8 m/s
v = 0 when 2t = 0 ⇒ t = 0; particle is stationary at t = 0.