Derivatives

Problem. Differentiate (a) y = \tan^{-1}\!\left(\dfrac{2x}{1-x^{2}}\right), (b) y = \tan^{-1}\sqrt{\dfrac{1-\cos x}{1+\cos x}}.

Solution. Both look terrible and both collapse under a substitution — always try that before the chain rule.

(a) Put x = \tan\theta, so \theta = \tan^{-1}x. Then \dfrac{2x}{1-x^{2}} = \dfrac{2\tan\theta}{1-\tan^{2}\theta} = \tan 2\theta, using the double-angle identity. Hence

(b) Use 1 - \cos x = 2\sin^{2}\frac{x}{2} and 1 + \cos x = 2\cos^{2}\frac{x}{2}:

Problem. Differentiate (a) y = e^{2x}\sin 3x, (b) y = \ln\left[\tan\!\left(\dfrac{x}{2} + \dfrac{\pi}{4}\right)\right].

Solution.

(a) Product rule, with each factor needing the chain rule:

(b) Let u = \tan\left(\frac{x}{2}+\frac{\pi}{4}\right). Then \dfrac{dy}{dx} = \dfrac{1}{u}\dfrac{du}{dx} and \dfrac{du}{dx} = \frac12\sec^{2}\left(\frac{x}{2}+\frac{\pi}{4}\right):

because \dfrac{\sec^{2}A}{\tan A} = \dfrac{1}{\sin A\cos A}. Now 2\sin A\cos A = \sin 2A with 2A = x + \frac{\pi}{2}:

Problem. If x^{3} + y^{3} = 3axy (the folium of Descartes), find \dfrac{dy}{dx}.

Solution. Differentiate every term with respect to x, remembering that y is a function of x so each y contributes a factor \dfrac{dy}{dx}:

Divide by 3 and collect the \dfrac{dy}{dx} terms on one side:

Problem. If x = a(\theta - \sin\theta) and y = a(1 - \cos\theta), find \dfrac{dy}{dx}.

Solution. Differentiate each coordinate with respect to the parameter and divide:

Use \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} and 1-\cos\theta = 2\sin^{2}\frac{\theta}{2}:

Problem. Differentiate y = x^{\sin x} + (\sin x)^{x}.

Solution. A variable raised to a variable power cannot be handled by the power rule or by the exponential rule. Split y = u + v and take logarithms of each part separately.

Let u = x^{\sin x}. Then \ln u = \sin x \ln x. Differentiating both sides:

Let v = (\sin x)^{x}. Then \ln v = x\ln(\sin x), so

Problem. If y = e^{m\sin^{-1}x}, prove that (1-x^{2})y_{2} - x y_{1} - m^{2}y = 0, where y_1 = \dfrac{dy}{dx} and y_2 = \dfrac{d^{2}y}{dx^{2}}.

Solution. Differentiate once, by the chain rule:

Clear the surd before differentiating again — this is the standard trick:

Differentiate this with respect to x:

Divide throughout by 2y_{1} (non-zero except at isolated points):

Problem. Differentiate (a) y = \cosh(x^{2}+1), (b) y = \ln(\cosh x), (c) y = \sinh^{3}2x, (d) y = \sinh x\cosh x.

Solution.

(a) Chain rule with u = x^{2}+1:

(b) \dfrac{dy}{dx} = \dfrac{1}{\cosh x}\cdot\sinh x = \tanh x.

(c) Write y = (\sinh 2x)^{3} and apply the chain rule twice:

(d) Product rule:

(the last step by the identity \cosh 2x = \cosh^2 x + \sinh^2 x; you could also write y = \frac12\sinh 2x first and differentiate in one line).

Problem. Differentiate (a) y = \sinh^{-1}\dfrac{x}{3}, (b) y = \cosh^{-1}(2x), (c) y = \tanh^{-1}(x^{2}).

Solution. Each is a standard derivative composed with an inner function.

(a) With u = \dfrac{x}{3}, \dfrac{du}{dx} = \dfrac13:

(b) With u = 2x:

(c) With u = x^{2}:

Problem. Evaluate (a) \displaystyle\lim_{x\to0}\frac{e^{x} - e^{-x} - 2x}{x - \sin x}, (b) \displaystyle\lim_{x\to0}\frac{\tan x - x}{x - \sin x}.

Solution.

(a) At x = 0 the numerator is 1 - 1 - 0 = 0 and the denominator is 0 - 0 = 0: the form is \frac00, so the rule applies.

(b) Again \frac00 at every stage:

using \tan x = \dfrac{\sin x}{\cos x} to cancel the \sin x at the third step.

Problem. Evaluate (a) \displaystyle\lim_{x\to\infty}\frac{x^{2}}{e^{x}}, (b) \displaystyle\lim_{x\to 0^{+}} x\ln x, (c) \displaystyle\lim_{x\to 0^{+}} x^{x}.

Solution.

(a) The form is \dfrac{\infty}{\infty}. Apply the rule twice:

The exponential beats every power — this is why e^{x} dominates in growth comparisons.

(b) The form is 0 \cdot (-\infty), which is indeterminate but not a quotient. Rewrite it as one:

(c) The form is 0^{0}. Take logarithms: let y = x^{x}, so \ln y = x\ln x. By part (b), \ln y \to 0, hence

Problem. Using L'Hospital's rule, evaluate (a) \displaystyle\lim_{x\to0}\frac{a^{x}-b^{x}}{x} (a, b > 0), (b) \displaystyle\lim_{x\to0}\frac{\sinh x}{x}, (c) \displaystyle\lim_{x\to0}\left(\frac1x - \frac{1}{\sin x}\right).

Solution.

(a) At x=0 the numerator is 1 - 1 = 0: the form is \frac00. Using \dfrac{d}{dx}a^{x} = a^{x}\ln a,

(b) Again \frac00:

the hyperbolic twin of \lim_{x\to0}\frac{\sin x}{x} = 1.

(c) The form is \infty - \infty, so combine into a single fraction first:

Problem. (a) Find the approximate value of \sqrt{49.5}. (b) The radius of a sphere is measured as 7 cm with a possible error of 0.02 cm. Find the approximate error in the calculated volume.

Solution.

(a) Take y = \sqrt{x} with x = 49 (a perfect square nearby) and \Delta x = 0.5. Then f'(x) = \dfrac{1}{2\sqrt{x}}, so

(The calculator value is 7.03562, so the estimate is correct to four decimal places.)

(b) V = \dfrac43\pi r^{3}, so \dfrac{dV}{dr} = 4\pi r^{2} and

The relative error is worth noting: \dfrac{dV}{V} = \dfrac{4\pi r^{2}dr}{\frac43\pi r^{3}} = 3\dfrac{dr}{r}, so a 0.29\% error in the radius becomes a 0.86\% error in the volume — errors are tripled by cubing.

Problem. The side of a cube is measured as 10 cm with a possible error of 0.01 cm. Find the approximate error in the calculated surface area, and the percentage error in the calculated volume.

Solution. Let a be the side, S = 6a^{2} the surface area and V = a^{3} the volume, with a = 10 and da = 0.01.

For the volume, use the relative-error shortcut. The percentage error in the side is

and since V = a^{3},

Problem. Find the equations of the tangent and the normal to the curve y = x^{3} - 3x + 2 at the point (2, 4).

Solution. First check the point lies on the curve: 2^{3} - 3(2) + 2 = 8 - 6 + 2 = 4 ✓.

Tangent: y - 4 = 9(x - 2), that is

Normal: slope = -\dfrac19, so y - 4 = -\dfrac19(x-2). Multiply by 9:

Problem. (a) Find the point on the curve y = x^{3} - 11x + 5 at which the tangent is the line y = x - 11. (b) Find the equation of the tangent to the circle x^{2} + y^{2} = 25 at (3, 4).

Solution.

(a) The given line has slope 1, so we need \dfrac{dy}{dx} = 3x^{2} - 11 = 1, giving x^{2} = 4, x = \pm 2.

• At x = 2: y = 8 - 22 + 5 = -9. The line at x=2 gives y = 2 - 11 = -9 ✓ — the point (2, -9) is on both.
• At x = -2: y = -8 + 22 + 5 = 19, but the line gives y = -13 ✗ — here the tangent is parallel to the line, not the line itself.

Hence the required point is (2, -9).

(b) Differentiate implicitly:

(Consistent with the Unit 8 result that the tangent to x^2+y^2=a^2 at (x_1,y_1) is xx_{1} + yy_{1} = a^{2}.)

Problem. Find the angle of intersection of the curves y = x^{2} and y^{2} = x at the point other than the origin.

Solution. Points of intersection. Substitute y = x^{2} into y^{2} = x:

So the required point is (1, 1).

Slopes there. For y = x^{2}: \dfrac{dy}{dx} = 2x, so m_{1} = 2.

For y^{2} = x, differentiate implicitly: 2y\dfrac{dy}{dx} = 1, so \dfrac{dy}{dx} = \dfrac{1}{2y} and m_{2} = \dfrac12.

Angle.

Problem. Air is pumped into a spherical balloon at 100 cm³ s⁻¹. Find the rate at which the radius is increasing when the radius is 15 cm. Find also the rate of increase of the surface area at that instant.

Solution. Let V, S and r be the volume, surface area and radius at time t. Then V = \dfrac43\pi r^{3} and S = 4\pi r^{2}.

Differentiate V with respect to t (chain rule):

Substitute \dfrac{dV}{dt} = 100 and r = 15:

For the surface area,

Problem. A ladder 5 m long rests against a vertical wall. The foot of the ladder is pulled away from the wall at 2 m s⁻¹. How fast is the top sliding down the wall when the foot is 4 m from the wall?

Solution. Let x be the distance of the foot from the wall and y the height of the top, at time t. By Pythagoras the ladder length is constant:

Differentiate with respect to t:

When x = 4: y = \sqrt{25 - 16} = 3, and \dfrac{dx}{dt} = 2:

The negative sign means y is decreasing: the top slides down at \dfrac83 m s⁻¹.

Problem. Water is poured into an inverted right circular cone at the rate of 3 m³ min⁻¹. The cone is 10 m deep and its base radius is 5 m. Find the rate at which the water level is rising when the water is 4 m deep.

Solution. Let h be the depth of water and r the radius of the water surface at time t. By similar triangles (the water cone is similar to the whole cone),

This is the constraint that removes r. The volume of water is

Differentiate with respect to t:

Now substitute \dfrac{dV}{dt} = 3 and h = 4:

Note that \dfrac{dh}{dt} is proportional to \dfrac{1}{h^{2}}: the deeper the water, the more slowly the level rises.

Problem. A particle moves along a straight line so that its displacement after t seconds is s = t^{3} - 6t^{2} + 9t + 4 metres. Find (a) its initial velocity, (b) the times when it is momentarily at rest and its displacement then, (c) its acceleration at those times.

Solution.

(a) Initial velocity is v at t = 0: v = 9 m s⁻¹.

(b) At rest when v = 0, that is t = 1 s and t = 3 s.

(c) a(1) = 6 - 12 = -6 m s⁻² and a(3) = 18 - 12 = 6 m s⁻².

Between t = 1 and t = 3 the velocity is negative, so the particle moves backwards by 8 - 4 = 4 m before turning round again.

Problem. Find the intervals in which f(x) = 2x^{3} - 9x^{2} + 12x + 5 is increasing and decreasing.

Solution.

f'(x) = 0 at x = 1 and x = 2. These split the real line into three intervals; test the sign of each factor in each.

So f is increasing on (-\infty, 1) and (2, \infty), and decreasing on (1, 2).

Problem. Find the local maximum and minimum values of f(x) = x^{3} - 6x^{2} + 9x + 15.

Solution.

At x = 1: f''(1) = 6 - 12 = -6 < 0, so there is a local maximum, of value

At x = 3: f''(3) = 18 - 12 = 6 > 0, so there is a local minimum, of value

Note the "maximum" 19 is larger than the "minimum" 15 only locally — the cubic itself is unbounded in both directions.

Problem. A square sheet of tin of side 18 cm has a small square cut from each corner, and the flaps are folded up to make an open box. Find the side of the square cut off so that the volume of the box is a maximum, and find that maximum volume.

Solution. Let x cm be the side of the square cut from each corner, where 0 < x < 9. The base is a square of side 18 - 2x and the height is x, so

Differentiate by the product rule:

Set \dfrac{dV}{dx} = 0: x = 9 or x = 3. Reject x = 9 (then the base has side 0 and V = 0), leaving x = 3.

At x = 3: \dfrac{d^{2}V}{dx^{2}} = 72 - 144 = -72 < 0, confirming a maximum. Hence

The square cut from each corner should have side 3 cm.

Problem. Show that the cylinder of greatest volume that can be inscribed in a sphere of radius R has height \dfrac{2R}{\sqrt{3}}, and find that volume.

Solution. Let the cylinder have radius r and height h. Its axis passes through the centre of the sphere, so a radius of the sphere, half the height, and the cylinder's radius form a right triangle:

This is the constraint that removes one variable. The volume is

Now R is a constant and h is the only variable:

so this is a maximum. The volume is

which is \dfrac{4\sqrt{3}\,\pi R^{3}}{9} \approx 0.7698 R^{3}, about 58\% of the sphere's own volume \frac43\pi R^{3}.

Problem. Find the absolute maximum and minimum values of f(x) = 2x^{3} - 24x + 107 on the interval [1, 3].

Solution.

Only x = 2 lies in [1, 3]; discard x = -2. Now evaluate f at that point and at both endpoints:

The absolute maximum is 89, at x = 3, and the absolute minimum is 75, at x = 2. Notice that the maximum occurs at an endpoint, where f'(x) \ne 0 — which is why endpoints must always be tested.

Problem. For f(x) = x^{4} - 4x^{3} + 10, find the intervals of concavity, the points of inflection, and the local extrema.

Solution.

Stationary points. f'(x) = 0 at x = 0 and x = 3.

• At x = 3: f''(3) = 12(3)(1) = 36 > 0, so a local minimum, value f(3) = 81 - 108 + 10 = -17.
• At x = 0: f''(0) = 0, so the second derivative test fails. Use the first derivative test: f'(x) = 4x^{2}(x-3) is negative on both sides of 0 (because 4x^{2} > 0 and x - 3 < 0), so f' does not change sign. Hence (0, 10) is not an extremum.

Concavity. f''(x) = 12x(x-2) = 0 at x = 0 and x = 2.

f'' changes sign at both x = 0 and x = 2, so there are two points of inflection: (0, 10) and (2, -6), since f(2) = 16 - 32 + 10 = -6.