Differentiating Exponential, Logarithmic and Trigonometric Functions — Answers
Answer sheet with worked solutions for the 12th Grade calculus worksheet.
Success TutoringWorksheetsAnswer sheet12th Grade · Mathematics · CalculusDifferentiating Exponential, Logarithmic and Trigonometric Functions — Answers
Part A: Standard derivatives
[frac d/dx] e[sup x] = e[sup x], [frac d/dx] ln x = [frac 1/x], [frac d/dx] sin x = cos x, [frac d/dx] cos x = −sin x, [frac d/dx] tan x = sec[sup 2] x. Angles are always in radians in calculus.
With the chain rule: ef(x) = f′(x) ef(x), ln f(x) = , sin f(x) = f′(x) cos f(x) and cos f(x) = −f′(x) sin f(x).
Worked example. y = e3x². Here f(x) = 3x², so = 6x e[sup 3x²]. Similarly ln(x2 + 1) = .
Answer: 4e4x
- Chain rule: ef(x) = f′(x) ef(x), with f(x) = 4x
- f′(x) = 4e[sup 4x]
Answer:
- =
- At x = 1: = [frac 3/4] (0.75)
- A−3 sin 3x
- B3 sin 3x
- C−sin 3x
- D−3 cos 3x
Answer: −3 sin 3x
cos x = −sin x, and the chain rule multiplies by the derivative of 3x, which is 3: −3 sin 3x.
Answer: 2
tan x = sec2 x = ; the chain rule multiplies by 2.
Answer: False
By the chain rule ln 5x = = . (Or: ln 5x = ln 5 + ln x, and ln 5 is a constant.)
Part B: Combining the rules
Product rule: (uv)′ = u′v + uv′. Quotient rule: ()′ = . Chain rule: = × .
Decide the outer structure first (a product, a quotient or a function of a function), then differentiate each piece with the standard results.
Worked example. y = x2 e−x. Product rule with u = x2, v = e−x: = 2x e−x − x2 e−x = x e[sup −x](2 − x).
Answer: 2
- Product rule: = 1 × ln x + x × = ln x + 1
- At x = e: ln e + 1 = 1 + 1 = 2
Answer: 1
- Product rule: = 2e2x sin x + e2x cos x
- At x = 0: 2 × 1 × 0 + 1 × 1 = 1
Answer: e
- Quotient rule: = =
- = 0 when 1 − ln x = 0, so ln x = 1 and x = e
Answer: 1
- Chain rule with y = (sin x)2: = 2 sin x cos x
- At x = : 2 × × = 2 × = 1
Answer:
- =
- At x = 2: = [frac 1/2]
Answer: −0.736
- Chain rule: f′(x) = −2x e−x²
- f′(1) = −2e−1 = − = −0.7357…
- f′(1) = −[frac 2/e] ≈ −0.736
Part C: Tangents, stationary points and rates of change
The slope of the tangent at x = a is f′(a); the tangent is y − f(a) = f′(a)(x − a). Stationary points occur where f′(x) = 0; classify them with the second derivative (f″ < 0 maximum, f″ > 0 minimum) or a table of slopes.
If Q depends on time t, then is the rate of change of Q, with units of Q per unit of time.
Worked example. Stationary points of y = ex − 2x: y′ = ex − 2 = 0 gives x = ln 2, y = 2 − 2 ln 2. Since y″ = ex > 0, it is a minimum turning point.
Answer: 2, 1
At x = 0, y = e0 = 1. = 2e2x = 2 at x = 0. Tangent: y − 1 = 2(x − 0), so y = 2x + 1.
Answer:
- = , so the slope at x = e is [frac 1/e].
- The point of contact is (e, ln e) = (e, 1).
- Tangent: y − 1 = (x − e), so y = − 1 + 1 = , which passes through (0, 0).
Answer: 1, maximum
y′ = e−x − x e−x = e−x(1 − x) = 0 gives x = 1 (e−x is never 0). y″ = e−x(x − 2), so y″(1) = − < 0: a maximum turning point at (1, ).
Answer: ,
- = cos x − sin x = 0, so sin x = cos x and tan x = 1
- Tangent is positive in quadrants 1 and 3: x = [frac π/4], [frac 5π/4]
- (These give a maximum of √2 and a minimum of −√2.)
Answer: 81
- = 2000 × 0.03 e0.03t = 60 e0.03t
- At t = 10: 60 e0.3 = 60 × 1.34985… = 80.99…
- The population is growing by about 81 individuals per year.
Part D: Extended problem
The depth of water in a bay t hours after midnight is modeled by D(t) = 5 + 2 sin() m, for 0 ≤ t ≤ 24.
(a) Find the greatest and least depths, and the first time each occurs.
(b) Find D′(t).
(c) Find the rate at which the depth is changing at 2 am, correct to 3 decimal places. Is the water rising or falling?
(d) Find the rate of change at 8 am and interpret it.
(e) Find the greatest rate at which the water rises, and the first time this happens.
Model answer: (a) sin() ranges from −1 to 1, so the greatest depth is 5 + 2 = 7 m and the least is 5 − 2 = 3 m. The first maximum is when = , t = 3 (3 am); the first minimum when = , t = 9 (9 am). (b) D′(t) = [frac π/3] cos([frac πt/6]). (c) D′(2) = cos() = ≈ 0.524 m per hour (rising). (d) D′(8) = cos() = − ≈ −0.524 m per hour: the water is falling at about 0.524 m per hour. (e) The greatest value of cos is 1, so the fastest rise is ≈ 1.047 m per hour, first at t = 0 (midnight), when the depth is 5 m and rising.
Success criteria:
- (a) greatest depth 7 m at t = 3 and least depth 3 m at t = 9 (2 marks)
- (b) correct derivative using the chain rule (2 marks)
- (c) ≈ 0.524 m/h, rising (2 marks)
- (d) − ≈ −0.524 m/h, interpreted as falling (2 marks)
- (e) ≈ 1.047 m/h at t = 0 (2 marks)
Marking tips
- Give credit for correct method even when the final answer slips — the working shows where the thinking went right.
- For open-ended tasks, use the success criteria as a checklist rather than looking for one "right" answer.
- Celebrate what went well first, then pick one thing to work on next.