Questions on differentiating trig, exponentials and logs

Question 1

Differentiate y = sin 2x, and y = cos 3x.

Accepted 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.

Question 2

Differentiate y = e^5x, and y = ln x.

Accepted Answer

e^5x gives 5e^5x, which is k times itself. ln x gives 1/x. B1 B1, one for each derivative.

Question 3

Find the gradient of the curve y = sin x at the point where x = 1, giving your answer to 3 significant figures, and state the unit in which x must be measured for the derivative you have used to be valid.

Accepted Answer

dy/dx = cos x, so the gradient at x = 1 is cos 1 = 0.540 to 3 significant figures. B1. The standard derivative of sin x is cos x only when x is measured in radians. B1. A calculator left in degree mode returns cos 1° = 0.99985, which is the cosine of a quite different angle, and in that measure the derivative of sin x carries an extra factor of π/180 = 0.017453, because (sin h)/h then tends to π/180 rather than to 1.

Question 4

Find the exact gradient of the curve y = 2 cos 3x at the point where x = π/9.

Accepted Answer

The derivative of cos kx is −k sin kx, so dy/dx = 2 × (−3 sin 3x) = −6 sin 3x. M1 for the standard derivative with its factor of 3, A1 for −6 sin 3x. At x = π/9 the angle 3x is π/3, and sin (π/3) = √3/2, so the gradient is −6 × √3/2 = −3√3. A1. The inner coefficient 3 multiplies the 2 that was already in front, and the minus sign belongs to the derivative of cosine rather than to the arithmetic. Work out the angle 3x before taking its sine, and keep the surd: a decimal does not answer a question that asks for an exact gradient.

Question 5

Differentiate y = 2 sin 3x + 4 cos 2x.

Accepted 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.

Question 6

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.

Accepted 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. Working in degrees turns cos 2π into cos 6.28° and quietly ruins the answer.

Question 7

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.

Accepted 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.

Question 8

Differentiate y = e^2x − 4e^−x, and find the gradient of the curve at x = 0.

Accepted Answer

dy/dx = 2e^2x + 4e^−x. The −1 in the second exponent meets the −4 in front, so that term is positive. At x = 0 both exponentials equal 1, so the gradient is 2 + 4 = 6. M1 A1 for the derivative, A1 for the value.

Question 9

Differentiate y = tan 3x, and find the gradient of the curve at x = 0.

Accepted 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.

Question 10

Find an equation of the tangent to the curve y = ln x at the point where x = e, and show that this tangent passes through the origin.

Accepted Answer

The point of contact is (e, ln e) = (e, 1), since ln e = 1. B1 for the point. The derivative of ln x is 1/x, so the gradient there is 1/e. M1 for the derivative, A1 for the gradient 1/e. Point-gradient form gives y − 1 = (1/e)(x − e). M1. Expanding the bracket, y − 1 = x/e − 1, so the tangent is y = x/e, and substituting x = 0 gives y = 0, so the line passes through the origin. A1. The 1 that fixes the height and the 1 that comes out of (1/e) × e cancel exactly, which is why no constant term survives. The general tangent, at x = a, is y = x/a + ln a − 1, and it meets the y-axis at ln a − 1, so x = e is the only point of contact whose tangent goes through the origin.

Question 11

Given that y = 5 × 3^2x, find dy/dx, and find the value of x for which dy/dx = 90 ln 3.

Accepted Answer

The derivative of a^kx is ka^kx ln a, so with a = 3 and k = 2, dy/dx = 5 × 2 × 3^2x ln 3 = 10 × 3^2x ln 3. M1 for the rule with both the factor k and the factor ln a, A1 for the derivative. Setting that equal to 90 ln 3 and dividing both sides by ln 3, which is not zero, leaves 10 × 3^2x = 90, so 3^2x = 9 = 3². M1 for reaching one power of 3 on each side. Comparing the indices, 2x = 2, so x = 1. A1. The variable sits in the exponent here, so the power rule has no claim on it, and 3^2x does not differentiate to 2x × 3^2x−1.

Question 12

Given that y = ln x, use the fact that x = e^y to show that dy/dx = 1/x.

Accepted 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.

Question 13

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.

Accepted Answer

Write a^x = e^x ln a, using a = e^ln a. The exponent is now x times the constant ln a, so the chain rule gives ln a × e^x ln a = a^x ln a. For y = 2^x, dy/dx = 2^x ln 2, and at x = 3 the gradient is 8 ln 2 = 5.55. M1 for rewriting in base e, M1 for differentiating, A1 for the printed result, M1 for the substitution, A1 for 5.55. The power rule does not apply to a^x, where the variable is the exponent, so x a^x−1 is not its derivative.

Question 14

Show that if y = e^kx, where k is a constant, then d²y/dx² = k²y. Hence show that y = 3e^2x − 5e^−2x satisfies d²y/dx² = 4y.

Accepted Answer

For y = e^kx, dy/dx = ke^kx, and differentiating a second time gives d²y/dx² = k²e^kx. M1 for the first derivative. Since e^kx is y itself, that is k²y. A1. Now apply it term by term, with k = 2 for the first term and k = −2 for the second. The first gives 3 × 2²e^2x = 12e^2x, and the second gives −5 × (−2)²e^−2x = −20e^−2x. M1 for differentiating the combination twice, A1 for d²y/dx² = 12e^2x − 20e^−2x. Factorising, that is 4(3e^2x − 5e^−2x) = 4y, as required. A1. Both exponents square to the same 4, so both terms are multiplied by the same factor and the combination survives intact. A term in e^3x added to y would be multiplied by 9 instead, and the equation would fail.

Question 15

The curves y = 2^x and y = x² meet at the point where x = 2 and at the point where x = 4. Find the gradient of each curve at each of these two points, giving the gradients of y = 2^x to 3 significant figures, and state at each point which of the two curves is the steeper.

Accepted Answer

The derivative of 2^x is 2^x ln 2, and the derivative of x² is 2x. B1 for both derivatives. At x = 2 the curves do meet, since 2² = 4 = 2², and the gradients are 4 ln 2 = 2.77 and 4. M1 for substituting x = 2, A1 for both values. Since 2.77 < 4, y = x² is the steeper there. A1. At x = 4 they meet again, since 2⁴ = 16 = 4², and the gradients are 16 ln 2 = 11.1 and 8. M1 for substituting x = 4. Since 11.1 > 8, y = 2^x is the steeper there. A1 for both values with the comparison. Between the two crossings the exponential changes from the flatter curve to the steeper one, which is how it can drop below the parabola at one meeting and climb back to it at the other. The factor ln 2 = 0.693 is the same at both points; what changes is 2^x itself, and beyond x = 4 the exponential stays the steeper of the two.