Questions on calculus with hyperbolic functions

Question 1

Write down the derivatives of sinh x and cosh x, and note the difference from the trig pattern.

Accepted Answer

sinh' = cosh and cosh' = sinh. B1 B1 for the two derivatives. Unlike cos, whose derivative is −sin, cosh differentiates with no minus anywhere.

Question 2

Differentiate y = cosh 3x, and find the gradient at x = 0.

Accepted Answer

Chain rule gives dy/dx = 3 sinh 3x. At x = 0, sinh 0 = 0, so the gradient is 0. M1 A1 for the derivative, A1 for the gradient. The curve has its minimum there, of value cosh 0 = 1.

Question 3

Differentiate y = sinh(2x − 1).

Accepted Answer

Chain rule: the derivative of the inner function 2x − 1 is 2, so dy/dx = 2 cosh(2x − 1). M1 for the chain rule set up, A1 for the factor 2, A1 for cosh(2x − 1).

Question 4

Write down ∫ cosh x dx and ∫ sech²x dx.

Accepted Answer

∫cosh x dx = sinh x + c and ∫sech²x dx = tanh x + c. B1 for the first, B1 for the second. Both are the reverse of a derivative stated earlier in this lesson, with no sign change to track.

Question 5

Find the equation of the tangent to y = cosh x at x = ln 2.

Accepted Answer

Gradient sinh(ln 2) = 3/4 and value cosh(ln 2) = 5/4, so y = 5/4 + (3/4)(x − ln 2). M1 A1 for the gradient, B1 for the value of y, A1 for the tangent. Exact hyperbolic values at logarithmic points come straight from the definitions, with no calculator required.

Question 6

Evaluate ∫ sinh x dx from 0 to ln 3, exactly.

Accepted Answer

∫sinh = cosh, both signs positive, so the value is [cosh x] from 0 to ln 3 = cosh(ln 3) − 1 = 5/3 − 1 = 2/3. The value cosh(ln 3) = 5/3 comes from (3 + 1/3)/2. M1 for the antiderivative, M1 for cosh(ln 3) = 5/3, A1 for 2/3.

Question 7

Evaluate ∫ 1/√(x² − 9) dx from 3 to 5, giving the answer as a single logarithm.

Accepted Answer

With a = 3 the integral is arcosh(x/3) from 3 to 5, that is arcosh(5/3) − arcosh 1. In log form this is ln(5/3 + √(25/9 − 1)) = ln(5/3 + 4/3) = ln 3, and arcosh 1 = 0. The answer is ln 3. M1 A1 for the arcosh form, M1 for the logarithmic form, A1 for ln 3.

Question 8

Using the identity cosh 2x = 2 cosh²x − 1, evaluate ∫ cosh²x dx from 0 to 1, exactly.

Accepted Answer

cosh²x = (1 + cosh 2x)/2, so the integral is [x/2 + sinh 2x/4] from 0 to 1 = 1/2 + sinh 2/4. Since sinh 2 = 2 sinh 1 cosh 1, this is (1 + sinh 1 cosh 1)/2 ≈ 1.407. M1 for rearranging the identity, M1 for integrating, A1 for the antiderivative, A1 for the exact value. This form of the identity matches cos 2x = 2 cos²x − 1; the sign change appears only in cosh 2x = 1 + 2 sinh²x.

Question 9

Let y = arsinh x, so sinh y = x. By differentiating implicitly, show that dy/dx = 1/√(x² + 1), and state the values of x for which this holds.

Accepted Answer

Differentiating sinh y = x with respect to x: cosh y (dy/dx) = 1, so dy/dx = 1/cosh y. M1 for the implicit differentiation. Since cosh²y − sinh²y = 1, cosh y = √(1 + sinh²y) = √(1 + x²), taking the positive root because cosh y > 0 for every real y. M1 for cosh y in terms of x, A1 for the positive root justified. So dy/dx = 1/√(x² + 1), valid for every real x, since sinh y = x has a solution y for every real x. A1 for the derivative stated with its domain.

Question 10

State the substitution that would remove the square root from ∫ 1/√(9x² + 4) dx, and rewrite the integral in terms of u before integrating.

Accepted Answer

Factor out the 9 first: 9x² + 4 = 9(x² + 4/9), so this is the x² + a² pattern with a = 2/3. B1 for identifying a = 2/3 from the factored form. The substitution is x = (2/3) sinh u, since then 9x² + 4 = 4 sinh²u + 4 = 4cosh²u, a perfect square under the root. M1 for the substitution. Then dx = (2/3) cosh u du and √(9x² + 4) = 2 cosh u, so the integral becomes ∫ (2/3) cosh u / (2 cosh u) du = ∫ 1/3 du. M1 for dx and the simplified root, A1 for the integral in u.

Question 11

Differentiate y = x sinh²x.

Accepted Answer

Product rule, with sinh²x differentiated by the chain rule: dy/dx = sinh²x + x × 2 sinh x cosh x = sinh²x + 2x sinh x cosh x. M1 for the product rule set up, M1 for the chain rule on sinh²x, A1 for 2 sinh x cosh x, A1 for the complete derivative. Using sinh 2x = 2 sinh x cosh x, this can also be written sinh²x + x sinh 2x, a shorter form for the same function.

Question 12

Using the substitution x = 2 sinh u, show that ∫ √(x² + 4) dx from 0 to 2√3 = 2 ln(2 + √3) + 4√3.

Accepted Answer

dx = 2 cosh u du and x² + 4 = 4 sinh²u + 4 = 4 cosh²u, so √(x² + 4) = 2 cosh u and the integral becomes ∫4 cosh²u du. The limits transform too, with x = 0 giving u = 0 and x = 2√3 giving sinh u = √3, so u = arsinh √3 = ln(√3 + 2). Using cosh²u = (1 + cosh 2u)/2 turns the integral into 2[u + sinh u cosh u]. At the upper limit sinh u = √3 and cosh u = 2, giving 2[ln(2 + √3) + 2√3] = 2 ln(2 + √3) + 4√3, about 9.562. M1 for dx = 2 cosh u du, M1 A1 for reducing the root to 2 cosh u, B1 for the transformed limits, M1 for the double angle identity, A1 for the antiderivative, A1 for the printed result. The two traps are leaving the limits in terms of x and forgetting that cosh u is the positive root throughout.

Question 13

Let y = arcosh x for x > 1 and z = artanh x for |x| < 1. By differentiating cosh y = x and tanh z = x implicitly, show that dy/dx = 1/√(x² − 1) and dz/dx = 1/(1 − x²), explaining why the positive square root is taken for dy/dx.

Accepted Answer

Differentiating cosh y = x: sinh y (dy/dx) = 1, so dy/dx = 1/sinh y. M1 for the implicit differentiation. Since cosh²y − sinh²y = 1, sinh y = √(cosh²y − 1) = √(x² − 1). M1 for sinh y in terms of x. The positive root is correct here because arcosh is defined by restricting cosh to y ≥ 0, the branch left over once the restriction that makes cosh one-to-one is applied, and sinh y ≥ 0 throughout that range. A1 for the positive root justified from the range of arcosh. So dy/dx = 1/√(x² − 1), for x > 1. Differentiating tanh z = x: sech²z (dz/dx) = 1, so dz/dx = 1/sech²z = 1/(1 − tanh²z) = 1/(1 − x²), using 1 − tanh²z = sech²z. M1 for the implicit differentiation and the identity, A1 for expressing sech²z in terms of x. So dz/dx = 1/(1 − x²), for |x| < 1. A1 for the derivative with its domain. Unlike arcosh, artanh needs no choice of sign, since sech²z is positive by definition regardless of the sign of tanh z.

Question 14

Evaluate ∫ √(x² − 16) dx from 5 to 8, choosing a substitution of your own and stating why you chose it, giving your answer to 4 decimal places.

Accepted Answer

The integrand contains x² − a² with a = 4, so the substitution is x = 4 cosh u, with u ≥ 0: then x² − 16 = 16 cosh²u − 16 = 16 sinh²u, removing the square root. B1 for identifying the x² − a² pattern with a = 4 and choosing a substitution that removes the root (x = 4 cosh u, or x = 4 sec θ). Then dx = 4 sinh u du and √(x² − 16) = 4 sinh u, so the integral becomes ∫16 sinh²u du = 8∫(cosh 2u − 1) du = 8[½sinh 2u − u] = 8 sinh u cosh u − 8u. M1 for dx and the reduced root, M1 for the double angle identity on sinh²u, A1 for the antiderivative in u. Converting back, sinh u = √(x² − 16)/4 and cosh u = x/4, so 8 sinh u cosh u = 8 × √(x² − 16)/4 × x/4 = x√(x² − 16)/2, and u = arcosh(x/4), giving the antiderivative x√(x² − 16)/2 − 8 arcosh(x/4). M1 for converting back to x. At x = 8: 8√48/2 − 8 arcosh 2 = 16√3 − 8 ln(2 + √3). At x = 5: 5(3)/2 − 8 arcosh(5/4) = 7.5 − 8 ln 2. The definite integral is 15.2223. A1 for the value. A sine substitution here would leave a negative quantity under a square root, since x² > 16 throughout the interval; the x² − a² pattern is what points to cosh rather than sin or tan.

Question 15

Evaluate ∫ 1/√(x² + 6x + 13) dx from −3 to 0, giving your answer in exact logarithmic form.

Accepted Answer

The quadratic under the root is not in the standard form, so complete the square first: x² + 6x + 13 = (x + 3)² + 4. M1 for completing the square. Substituting u = x + 3, du = dx, and the limits x = −3 and x = 0 become u = 0 and u = 3: ∫1/√(u² + 4) du = arsinh(u/2), a standard form with a = 2. M1 for the substitution u = x + 3, A1 for recognising the standard arsinh form. So the integral is arsinh(3/2) − arsinh(0) = arsinh(3/2), since arsinh(0) = 0. M1 for evaluating at the transformed limits. In logarithmic form, arsinh(3/2) = ln(3/2 + √(9/4 + 1)) = ln(3/2 + √13/2) = ln((3 + √13)/2), about 1.1948. A1 A1 for the logarithmic form and the simplified argument.