HL · Further differentiation 1 · Derivatives of new functions

Module 38 of 47Lesson 1 of 8 · HL · Further differentiation 1 · Derivatives of new functions

Further differentiation · Essential understanding 1

Derivatives of new functions

IB AI AHL 5.9with Prof. Kitso

💻 Using the TI-Nspire CX Student Software on a laptop

The screens in these notes are drawn exactly as the Student Software shows them. The keypad on the left of the software works like the handheld — click its keys with the mouse.
menu: click menu on the on-screen keypad (or Documents Toolbox ▸ 🔧). The menu numbers and letters are the same as in these notes.
enter = Enter · tab = Tab (next box) · esc = Esc · del = Backspace.
Powers: type ^ (Shift+6), the power, then press → to come back down before typing more. x² = ^2 then →.
Fraction: type / · √: type sqrt( · π: type pi · eˣ: type exp( · store: type :=. For EE (×10ⁿ) and i, click them on the keypad.
New page: Ctrl+I (or the Insert menu) · change page: Ctrl+← / Ctrl+→ · graph entry line: Ctrl+G · var: Ctrl+L.
Radian / Degree: click Settings in the status bar at the bottom of the window. The mode (RAD or DEG) shows in the top bar.
IB exam mode (Press-to-Test): everything in these notes works on the non-CAS TI-Nspire with the IB exam settings. In exam mode there are no dot-product, cross-product, unit-vector or eigenvalue keys and no Diff Eq graphs — the notes show the allowed way (type the components, use polyRoots on the characteristic equation, do Euler steps on a Calculator page).
If you type nDeriv(, the software shows it as centralDiff( — it is the same command.

The big idea: Until now you could only differentiate powers like x², x⁵ and

\dfrac{1}{x}

. In this lesson your list grows. You will learn the derivatives of fractional powers (like √x), of sin x, cos x and tan x, of eˣ and of ln x. Learn these six results well: every harder rule in this unit is built on top of them.

By the end of this lesson you can:

differentiate xⁿ when n is any fraction or negative number
write roots and reciprocals as powers before differentiating
differentiate sin x, cos x and tan x (with x in radians)
differentiate eˣ and ln x
differentiate sums of these functions and find the gradient at a point
find the equation of a tangent to a curve that uses these functions
find where the gradient is zero in an exam-style question

🧑🏾 Imagine this…

Picture the curve y = sin x as a road over rolling hills. Walk along it from left to right. At the bottom of a valley the road is flat, then it climbs, gets steepest as it crosses the middle line, then eases off at the top of the hill. If you wrote down the steepness of the road at every point and plotted those numbers, you would draw another wave of exactly the same shape, just shifted along. That second wave is cos x. So the gradient of sin x is cos x. The graph is telling you the rule.

Derivative

the gradient function. For

y = f(x)

we write it

f'(x)

\dfrac{dy}{dx}

Rational power

a power that is a fraction, like

x^{\frac{1}{2}}

(that is √x) or

x^{-\frac{4}{3}}

. The symbol ℚ means “any fraction”.

Radians

the angle unit used in calculus. All the trigonometry results below are only true when x is in radians.

the special number

e = 2.71828\ldots

. The curve

y = e^x

has gradient equal to its own height at every point.

★ The six derivatives to know

$f(x) = x^n$ gives $f'(x) = nx^{n-1}$ , for any rational n (positive, negative or a fraction).
$\dfrac{d}{dx}(\sin x) = \cos x$
$\dfrac{d}{dx}(\cos x) = -\sin x$ (note the minus sign)
$\dfrac{d}{dx}(\tan x) = \dfrac{1}{\cos^2 x}$
$\dfrac{d}{dx}(e^x) = e^x$ (it does not change!)
$\dfrac{d}{dx}(\ln x) = \dfrac{1}{x}$ (for $x > 0$ )
A constant multiple stays in front, and sums and differences are differentiated term by term: (a f(x) + b g(x))′ = a f′(x) + b g′(x).
Rewrite first: $\sqrt{x} = x^{\frac{1}{2}}$ , $\sqrt[3]{x} = x^{\frac{1}{3}}$ , $\dfrac{1}{x^3} = x^{-3}$ , $\dfrac{1}{\sqrt{x}} = x^{-\frac{1}{2}}$ .

⚠️ Mistakes that lose marks

Forgetting to subtract 1 from a fraction. For $x^{\frac{1}{3}}$ the new power is $\dfrac{1}{3} - 1 = -\dfrac{2}{3}$ , not $\dfrac{1}{3}$ and not $\dfrac{2}{3}$ .
Degrees instead of radians. Check the top of the calculator screen says RAD. In degree mode the derivative of sin x is not cos x and your numbers will be wrong.
Sign of cos x. The derivative of cos x is −sin x. The derivative of sin x is +cos x. A quick memory test: cos x starts at its top (x = 0) and goes down, so its gradient there is negative or zero.
Differentiating eˣ as if it were a power. $\dfrac{d}{dx}(e^x)$ is $e^x$ , not $xe^{x-1}$ . The variable is in the power, so the power rule does not apply.

💡 Command terms

Differentiate and find f′(x) mean the same thing: give the gradient function. Find the gradient at x = a means substitute a into f′(x) at the end. If a question says show that, you must show every line, including the rewriting step.

🧪 Try it yourselfWatch the gradient of sin x turn into cos x

The blue curve is f(x) = sin x and the red curve is cos x. Drag the tangent along the blue curve and read its gradient. At x = 1 the gradient is 0.540, which is exactly the height of the red curve there (cos 1 = 0.540). At the top of a hill the tangent is flat and the red curve crosses zero. Where is the tangent steepest?

Worked examples

Example 1.1Fractional and negative powers

Differentiate (a)

f(x) = 5\sqrt{x}

(b)

g(x) = \dfrac{12}{\sqrt[3]{x}}

(c)

h(x) = x^{\frac{3}{2}} + \dfrac{2}{x^2}

. Then find

f'(4)

g'(8)

and

h'(4)

🧠 Think it through

Rewrite every term as a number times a power of x. √x is x to the power one half. 12 over ∛x is 12 times x to the power minus one third. 2 over x² is 2x to the power −2.
Use the power rule on each term: multiply by the power, then subtract 1 from the power.
Substitute the x value last. Check with the GDC derivative template.

✍️ By hand

(a)

f(x) = 5x^{\frac{1}{2}}

→

f'(x) = 5 \times \dfrac{1}{2}x^{-\frac{1}{2}} = \dfrac{5}{2\sqrt{x}}

f'(4) = \dfrac{5}{2 \times 2} = 1.25

(b)

g(x) = 12x^{-\frac{1}{3}}

→

g'(x) = 12 \times \left(-\dfrac{1}{3}\right)x^{-\frac{4}{3}} = -4x^{-\frac{4}{3}}

g'(8) = -4 \times 8^{-\frac{4}{3}} = -4 \times \dfrac{1}{16} = -0.25

(c)

h(x) = x^{\frac{3}{2}} + 2x^{-2}

→

h'(x) = \dfrac{3}{2}x^{\frac{1}{2}} - 4x^{-3}

h'(4) = \dfrac{3}{2}(2) - \dfrac{4}{64} = 3 - 0.0625 = 2.9375

🔢 On your TI-Nspire CX II

Check on a Calculator page with the derivative template (menu💻 click menu on keypad ▸ 4: Calculus ▸ 1: Numerical Derivative at a Point).

✅ Answer (a)

f'(x) = \dfrac{5}{2\sqrt{x}}

, f′(4) = 1.25. (b)

g'(x) = -4x^{-\frac{4}{3}}

, g′(8) = −0.25. (c)

h'(x) = \dfrac{3}{2}\sqrt{x} - \dfrac{4}{x^3}

, h′(4) = 2.9375.

The GDC gives the number but not the formula. In a “find f′(x)” question the working must be written by hand.

Example 1.2Sin, cos and tan

Let

f(x) = 3\sin x - 2\cos x + \tan x

, with x in radians. (a) Find

f'(x)

. (b) Find

f'\!\left(\dfrac{\pi}{3}\right)

and

f'(0)

🧠 Think it through

Differentiate each term: sin x → cos x, cos x → −sin x, tan x → $\dfrac{1}{cos^{2}x}$ .
Keep the number in front of each term.
Take care with the signs: −2 × (−sin x) = +2 sin x.

✍️ By hand

(a)

f'(x) = 3\cos x - 2(-\sin x) + \dfrac{1}{\cos^2 x} = 3\cos x + 2\sin x + \dfrac{1}{\cos^2 x}

(b)

f'\!\left(\dfrac{\pi}{3}\right) = 3\left(\dfrac{1}{2}\right) + 2\left(\dfrac{\sqrt{3}}{2}\right) + \dfrac{1}{(\frac{1}{2})^2} = 1.5 + \sqrt{3} + 4 = 7.23

(3 s.f.).

f'(0) = 3(1) + 2(0) + \dfrac{1}{1^2} = 4

🔢 On your TI-Nspire CX II

Set the calculator to Radian mode (the top of the screen shows RAD). Use the derivative template.

✅ Answer (a)

f'(x) = 3\cos x + 2\sin x + \dfrac{1}{\cos^2 x}

. (b) f′(

\dfrac{\pi}{3}

) = 7.23 and f′(0) = 4.

You may also meet

\dfrac{1}{\cos^2 x}

written as

\sec^2 x

. They are the same thing.

Example 1.3eˣ, ln x and a tangent line

A curve has equation

y = 4e^x - 3\ln x + x^2

for

x > 0

. (a) Find

\dfrac{dy}{dx}

and the gradient at

x = 2

. (b) Find the equation of the tangent to

y = \ln x

at the point where

x = 2

🧠 Think it through

eˣ stays as eˣ, ln x becomes $\dfrac{1}{x}$ , x² becomes 2x.
For (b) the tangent has equation y − y₁ = m(x − x₁), with m = the derivative at x = 2 and y₁ = ln 2.
Leave ln 2 exact in the working. Convert at the end.

✍️ By hand

(a)

\dfrac{dy}{dx} = 4e^x - \dfrac{3}{x} + 2x

. At

x = 2

4e^2 - \dfrac{3}{2} + 4 = 32.1

(3 s.f.).

(b) For

y = \ln x

\dfrac{dy}{dx} = \dfrac{1}{x}

, so at

x = 2

the gradient is

\dfrac{1}{2}

and the point is

(2, \ln 2)

y - \ln 2 = \dfrac{1}{2}(x - 2)

→

y = \dfrac{1}{2}x + \ln 2 - 1

→

y = 0.5x - 0.307

🔢 On your TI-Nspire CX II

Calculator page: check the gradient in (a), then the numbers for (b).

✅ Answer (a)

\dfrac{dy}{dx} = 4e^x - \dfrac{3}{x} + 2x

; gradient 32.1. (b)

y = 0.5x - 0.307

ln 2 − 1 is a number (−0.307). It is the y-intercept of the tangent. Do not drop the “−1”.

Example 1.4Rewrite first, then differentiate

Let

f(x) = \dfrac{x^2 + 4x}{\sqrt{x}}

, for

x > 0

. (a) Show that

f(x) = x^{\frac{3}{2}} + 4x^{\frac{1}{2}}

. (b) Find

f'(x)

and

f'(4)

. (c) Find the values of x for which

f'(x) = 3.5

🧠 Think it through

Divide each term on top by √x, using the index law $x^a \div x^b = x^{a-b}$ .
Differentiate the two powers. Rewrite the answer with a positive denominator.
For (c) put u = √x to turn the equation into a quadratic.

✍️ By hand

(a)

\dfrac{x^2}{x^{\frac{1}{2}}} = x^{\frac{3}{2}}

and

\dfrac{4x}{x^{\frac{1}{2}}} = 4x^{\frac{1}{2}}

, so

f(x) = x^{\frac{3}{2}} + 4x^{\frac{1}{2}}

. ✓

(b)

f'(x) = \dfrac{3}{2}x^{\frac{1}{2}} + 2x^{-\frac{1}{2}} = \dfrac{3}{2}\sqrt{x} + \dfrac{2}{\sqrt{x}}

f'(4) = 3 + 1 = 4

(c)

\dfrac{3}{2}u + \dfrac{2}{u} = 3.5

with

u = \sqrt{x}

. Multiply by 2u:

3u^2 + 4 = 7u

→

3u^2 - 7u + 4 = 0

→

(3u - 4)(u - 1) = 0

u = 1

u = \dfrac{4}{3}

, giving

x = 1

x = \dfrac{16}{9}

🔢 On your TI-Nspire CX II

Check part (b), then solve part (c) with nSolve. Use two different starting guesses to find both answers.

✅ Answer (b)

f'(x) = \dfrac{3}{2}\sqrt{x} + \dfrac{2}{\sqrt{x}}

; f′(4) = 4. (c) x = 1 or x =

\dfrac{16}{9}

= 1.78.

A ‘hidden quadratic’ gives two answers. If nSolve only shows one, change the starting guess.

Example 1.5Where is the gradient zero?

A curve has equation

y = x + 2\cos x

for

0 \le x \le 2\pi

. (a) Find

\dfrac{dy}{dx}

. (b) Find the x-coordinates of the points where the gradient is zero. (c) Find the y-coordinates, and say which point is a maximum and which is a minimum.

🧠 Think it through

Differentiate, then set $\dfrac{dy}{dx} = 0$ . This gives sin x = ½.
Solve in radians for x between 0 and 2π. The sine is positive in the first and second quadrants, so there are two answers.
To decide max or min, test the sign of the gradient on either side (or look at the graph).

✍️ By hand

(a)

\dfrac{dy}{dx} = 1 - 2\sin x

(b)

1 - 2\sin x = 0

→

\sin x = \dfrac{1}{2}

→

x = \dfrac{\pi}{6}

x = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}

(c)

y\!\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{6} + 2\cos\dfrac{\pi}{6} = \dfrac{\pi}{6} + \sqrt{3} = 2.26

y\!\left(\dfrac{5\pi}{6}\right) = \dfrac{5\pi}{6} - \sqrt{3} = 0.886

Gradient signs: at

x = 0

it is

1 - 0 = 1 > 0

; at

x = \dfrac{\pi}{2}

it is

1 - 2 = -1 < 0

; at

x = \pi

it is

1 > 0

. Up, down, up: so

x = \dfrac{\pi}{6}

is a maximum and

x = \dfrac{5\pi}{6}

is a minimum.

🔢 On your TI-Nspire CX II

Graph f1(x)=x+2cos(x) in Radian mode with window 0 ≤ x ≤ 6.28. Use menu💻 click menu on keypad▸6: Analyze Graph▸3: Maximum and menu💻 click menu on keypad▸6: Analyze Graph▸2: Minimum.
Or solve $\dfrac{dy}{dx} = 0$ on a Calculator page.

✅ Answer (a)

1 - 2\sin x

. (b)

x = \dfrac{\pi}{6}

(0.524) and

x = \dfrac{5\pi}{6}

(2.62). (c) Maximum at (0.524, 2.26); minimum at (2.62, 0.886).

Leaving the first answer as

\dfrac{\pi}{6}

is fine, and it is exact. On the GDC 0.523599 =

\dfrac{\pi}{6}

Challenge questions

⭐ Challenge 1.6A stationary point with eˣPaper 1 style

Let

f(x) = e^x - 3x

. (a) Find

f'(x)

. (b) Find the exact value of x for which

f'(x) = 0

. (c) Find the y-coordinate of this point, to 3 s.f.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Differentiate: eˣ → eˣ, −3x → −3.
Solve eˣ = 3 by taking natural logs.
Substitute back into f(x), not into f′(x).

✍️ By hand

(a)

f'(x) = e^x - 3

(b)

e^x - 3 = 0

→

e^x = 3

→

x = \ln 3

(c)

f(\ln 3) = e^{\ln 3} - 3\ln 3 = 3 - 3\ln 3 = -0.296

🔢 On your TI-Nspire CX II

Check on a Calculator page.

✅ Answer (a)

e^x - 3

. (b)

x = \ln 3

. (c)

y = 3 - 3\ln 3 = -0.296

Because eˣ and ln x are inverses,

e^{\ln 3}

is just 3. This trick appears again and again.

⭐ Challenge 1.7Visitors to a lodgePaper 2 style

A new lodge opens in Kasane. The number of guests staying each night, t months after opening, is modelled by

N(t) = 50 + 12\sqrt{t}

, for

t \ge 1

(a) Find $N'(t)$ .
(b) Find the rate of change of guests per night when t = 9, and state its units.
(c) Find the value of t when the number of guests is increasing at 1 guest per night per month.
(d) Find N(36), and explain why the model cannot be trusted for very large t.
(e) Explain, using $N'(t)$ , why the number of guests keeps increasing but more and more slowly.

✏️ Try it on paper and on your calculator first. Then check.

Show the full solution

🧠 Think it through

Write √t as t^½ and differentiate.
Substitute t = 9. Units are “guests per night, per month”.
Solve N′(t) = 1 for t.
Look at what happens to N′(t) as t gets larger.

✍️ By hand
(a)  → .
(b)  guests per night per month.
(c)  →  → .
(d)  guests. The model keeps growing without limit, but a lodge only has a fixed number of rooms.
(e)  is always positive, so N always increases. As t grows,  grows, so N′(t) gets smaller: growth slows down.

🔢 On your TI-Nspire CX II

Check (b) and (c).

✅ Answer (a)

N'(t) = \dfrac{6}{\sqrt{t}}

. (b) 2 guests per night per month. (c) t = 36. (d) 122 guests; a lodge has a limited number of rooms. (e) N′(t) > 0 but decreases as t increases.

Next: Lesson 2 · The chain rule

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL HL · Further differentiation 1 · Derivatives of new functions

IB DP Mathematics: Applications and Interpretation (AI) — SL and HL

Derivatives of new functions

Worked examples

Challenge questions