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Differentiating trig, exponentials and logs questions

The power rule ran Year 12; now the rest of the standard functions join the calculus. The derivative of sine is cosine, the derivative of ex is itself, and the derivative of ln x is 1/x. Each of these results depends on the angle being measured in radians.

9 original questions · 29 marks · the differentiating trig, exponentials and logs notes · Differentiation

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

2 marks · warm-up
Differentiate y = sin 2x, and y = cos 3x.
Worked answer
sin 2x gives 2 cos 2x, and cos 3x gives −3 sin 3x. B1 B1, one for each derivative. The inner coefficient multiplies out to the front, and cosine's derivative carries the minus sign.
2 marks · warm-up
Differentiate y = e^5x, and y = ln x.
Worked answer
e^5x gives 5e^5x, which is k times itself. ln x gives 1/x. B1 B1, one for each derivative. The logarithm is the one entry on the shelf whose derivative leaves its own family entirely, and that is worth memorising rather than deriving under exam pressure.
3 marks
Differentiate y = 2 sin 3x + 4 cos 2x.
Worked answer
Work termwise. 2 sin 3x gives 6 cos 3x, and 4 cos 2x gives −8 sin 2x, so dy/dx = 6 cos 3x − 8 sin 2x. M1 for a correct attempt at either term, then A1 A1. Each term's inner coefficient multiplies its own front constant, so the 3 pairs with the 2 and the 2 pairs with the 4, never the other way about.
4 marks
Differentiate y = 5 ln x + sin 2x, and hence find the gradient of the curve at x = π, giving your answer to 3 significant figures.
Worked answer
dy/dx = 5/x + 2 cos 2x. At x = π, cos 2π = 1, so the gradient is 5/π + 2 = 3.59. M1 A1 for the derivative, M1 for the substitution, A1 for the value. The logarithm's derivative still has an x in it, so the gradient genuinely depends on where you measure it. Working in degrees turns cos 2π into cos 6.28° and quietly ruins the answer.
4 marks · prove
Prove from first principles that the derivative of sin x is cos x. You may use the results (cos h − 1)/h → 0 and (sin h)/h → 1 as h → 0.
Worked answer
The chord gradient is [sin (x + h) − sin x]/h. Expanding the compound angle gives [sin x cos h + cos x sin h − sin x]/h, and regrouping gives sin x (cos h − 1)/h + cos x (sin h)/h. Letting h → 0 and quoting the two given limits leaves 0 + cos x × 1 = cos x. M1 for the chord gradient, M1 for the compound-angle expansion, M1 for the regrouping, A1 for the conclusion. The proof needs radians throughout, since (sin h)/h → 1 is false in degrees.
3 marks
Differentiate y = e^2x − 4e^−x, and find the gradient of the curve at x = 0.
Worked answer
dy/dx = 2e^2x + 4e^−x. The −1 in the second exponent meets the −4 in front and the term turns positive, which is the sign most often lost. At x = 0 both exponentials equal 1, so the gradient is 2 + 4 = 6. M1 A1 for the derivative, A1 for the value.
3 marks
Differentiate y = tan 3x, and find the gradient of the curve at x = 0.
Worked answer
tan kx differentiates to k sec² kx, so dy/dx = 3 sec² 3x. At x = 0, sec 0 = 1 and the gradient is 3. M1 A1 A1. Since sec² is never less than 1, this gradient never drops below 3, so tan 3x has no stationary points anywhere.
3 marks · stretch
Given that y = ln x, use the fact that x = e^y to show that dy/dx = 1/x.
Worked answer
From x = e^y, differentiate with respect to y to get dx/dy = e^y. Inverting, dy/dx = 1/e^y, and since e^y = x this is 1/x as required. M1 for dx/dy, M1 for turning the rate upside down, A1 for replacing e^y by x. The final substitution is the mark students forget, leaving the answer in terms of y when the question asked for x.
5 marks · show · stretch
Show that the derivative of a^x, for a constant a > 0, is a^x ln a. Hence find the gradient of the curve y = 2^x at x = 3, giving your answer to 3 significant figures.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

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