Differentiating sin x, cos x, tan x, eˣ and ln x

Three situations where a wave, a cooling curve or a logarithmic fill rate decides what is measured. The pictures come first; the working in each solution comes from the same numbers.

$\text{high water }5\text{ m}$$\text{low water }1\text{ m}$$\text{side view}$

Maths In Real Life 1

The tide at a sea wall

Tides rise and fall in a smooth wave. Port engineers use the highest and lowest water levels to set the height of a sea wall, and the rate of rise tells them how quickly the water reaches a given mark.

The height of the water at a sea wall is $y$ metres above the low-tide datum, where $y = 2 \sin{\left(\frac{\pi x}{6} \right)} + 3$, and $x$ is the time in hours after midnight, for $0 \le x \le 12$. The angle is in radians.

• (a)Find the times at which the water is highest and lowest, the height of the water at each of those times, and say which is which.
• (b)Find how fast the water is rising at midnight, when $x = 0$.

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Colour keyderivative of a part or intermediate resultfinal result

(a)

$\displaystyle \frac{d}{dx}\left(3\right) = \textcolor{#010105}{0}$

The derivative of a constant is $0$.

$\displaystyle \frac{d}{dx}\left(2 \sin{\left(\frac{\pi x}{6} \right)}\right) = 2 \cos{\left(\frac{\pi x}{6} \right)} \times \frac{\pi}{6} = \textcolor{#010105}{\frac{\pi \cos{\left(\frac{\pi x}{6} \right)}}{3}}$

Chain rule: the derivative of $\sin{\left(\frac{\pi x}{6} \right)}$ keeps the whole argument, giving $\cos{\left(\frac{\pi x}{6} \right)}$, then multiply by the derivative of the argument, $\frac{\pi}{6}$. The constant multiplier stays in front.

$\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{\frac{\pi \cos{\left(\frac{\pi x}{6} \right)}}{3}}$

Add the derivatives of the terms.

At a stationary point the gradient is $0$, so set the derivative equal to $0$ for $0 < x < 12$.

$\displaystyle \frac{\pi \cos{\left(\frac{\pi x}{6} \right)}}{3} = 0$

A stationary point is where the gradient is zero.

$\displaystyle x\;= \textcolor{#010105}{3} \text{ or } x = \textcolor{#010105}{9}$

Solve for $x$ in radians, keeping only the values inside the interval.

$\displaystyle y\;= 2 \times \sin{\left(\frac{\pi \times \left(3\right)}{6} \right)} + 3 = \textcolor{#010106}{5}$

Put $x = 3$ into the curve.

$\displaystyle y\;= 2 \times \sin{\left(\frac{\pi \times \left(9\right)}{6} \right)} + 3 = \textcolor{#010106}{1}$

Put $x = 9$ into the curve.

Differentiate the first derivative once more.

$\displaystyle \frac{d^{2}y}{dx^{2}}\;= \left(- \frac{\pi \sin{\left(\frac{\pi x}{6} \right)}}{3}\right) \times \frac{\pi}{6}$

Chain rule: the derivative of $\cos{\left(\frac{\pi x}{6} \right)}$ keeps the whole argument, giving $- \sin{\left(\frac{\pi x}{6} \right)}$, then multiply by the derivative of the argument, $\frac{\pi}{6}$. The constant multiplier stays in front.

$\displaystyle = \textcolor{#010106}{- \frac{\pi^{2} \sin{\left(\frac{\pi x}{6} \right)}}{18}}$

Combine the factors into one term.

$\displaystyle y''\;= - \frac{\pi^{2} \times \sin{\left(\frac{\pi \times \left(3\right)}{6} \right)}}{18} = \textcolor{#010106}{- \frac{\pi^{2}}{18}}$

Put $x = 3$ into the second derivative. Use the exact trigonometric values; the angle is in radians.

The second derivative is negative, so the point is a maximum.

$\displaystyle f''\;= - \frac{\pi^{2} \times \sin{\left(\frac{\pi \times \left(9\right)}{6} \right)}}{18} = \textcolor{#010106}{\frac{\pi^{2}}{18}}$

Put $x = 9$ into the second derivative. Use the exact trigonometric values; the angle is in radians.

The second derivative is positive, so the point is a minimum.

(b)

$\displaystyle \frac{d}{dx}\left(3\right) = \textcolor{#010105}{0}$

The derivative of a constant is $0$.

$\displaystyle \frac{d}{dx}\left(2 \sin{\left(\frac{\pi x}{6} \right)}\right) = 2 \cos{\left(\frac{\pi x}{6} \right)} \times \frac{\pi}{6} = \textcolor{#010105}{\frac{\pi \cos{\left(\frac{\pi x}{6} \right)}}{3}}$

Chain rule: the derivative of $\sin{\left(\frac{\pi x}{6} \right)}$ keeps the whole argument, giving $\cos{\left(\frac{\pi x}{6} \right)}$, then multiply by the derivative of the argument, $\frac{\pi}{6}$. The constant multiplier stays in front.

$\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{\frac{\pi \cos{\left(\frac{\pi x}{6} \right)}}{3}}$

Add the derivatives of the terms.

$\displaystyle \frac{dy}{dx}\;= \frac{\pi \times \cos{\left(\frac{\pi \times \left(0\right)}{6} \right)}}{3} = \textcolor{#010106}{\frac{\pi}{3}}$

Put $x = 0$ into the derivative, then use the exact values of the trigonometric functions at this angle (the angle is in radians).

Answer

(a) $x = 3$ hours after midnight, $y = 5$ m; $x = 9$ hours after midnight, $y = 1$ m; high water: $y = 5$ m at $x = 3$ hours after midnight; low water: $y = 1$ m at $x = 9$ hours after midnight

(b) Rate (metres per hour): $\frac{dy}{dx} = \frac{\pi}{3}$ when $x = 0$

$\text{start }80\text{ °C}$$\text{front view}$

Maths In Real Life 2

A cup of tea cooling

A hot drink loses heat fastest at the start and more slowly as it nears room temperature. The gradient of the cooling curve measures how fast, and its sign tells you whether the cooling is speeding up or slowing down.

A cup of tea is poured at time $x = 0$. Its temperature is $y$ °C at $x$ minutes after pouring, where $y = 24 + 56 e^{- \frac{x}{20}}$.

• (a)Find the rate at which the temperature is changing at the instant the tea is poured, when $x = 0$.
• (b)Find $\frac{d^{2}y}{dx^{2}}$ and its value when $x = 0$, and say whether the cooling is speeding up or slowing down.

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Colour keyderivative of a part or intermediate resultfinal result

(a)

$\displaystyle \frac{d}{dx}\left(24\right) = \textcolor{#010105}{0}$

The derivative of a constant is $0$.

$\displaystyle \frac{d}{dx}\left(56 e^{- \frac{x}{20}}\right) = 56 e^{- \frac{x}{20}} \times \left(- \frac{1}{20}\right) = \textcolor{#010105}{- \frac{14 e^{- \frac{x}{20}}}{5}}$

Chain rule: the derivative of $e^{- \frac{x}{20}}$ keeps the whole argument, giving $e^{- \frac{x}{20}}$, then multiply by the derivative of the argument, $- \frac{1}{20}$. The constant multiplier stays in front.

$\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{- \frac{14 e^{- \frac{x}{20}}}{5}}$

Add the derivatives of the terms.

$\displaystyle \frac{dy}{dx}\;= - \frac{14 \times e^{- \frac{\left(0\right)}{20}}}{5} = \textcolor{#010106}{- \frac{14}{5}}$

Put $x = 0$ into the derivative, then work out the exact value.

(b)

$\displaystyle \frac{d}{dx}\left(24\right) = \textcolor{#010105}{0}$

The derivative of a constant is $0$.

$\displaystyle \frac{d}{dx}\left(56 e^{- \frac{x}{20}}\right) = 56 e^{- \frac{x}{20}} \times \left(- \frac{1}{20}\right) = \textcolor{#010105}{- \frac{14 e^{- \frac{x}{20}}}{5}}$

Chain rule: the derivative of $e^{- \frac{x}{20}}$ keeps the whole argument, giving $e^{- \frac{x}{20}}$, then multiply by the derivative of the argument, $- \frac{1}{20}$. The constant multiplier stays in front.

$\displaystyle \frac{dy}{dx}\;= \textcolor{#010106}{- \frac{14 e^{- \frac{x}{20}}}{5}}$

Add the derivatives of the terms.

Differentiate the first derivative once more.

$\displaystyle \frac{d^{2}y}{dx^{2}}\;= \left(- \frac{14 e^{- \frac{x}{20}}}{5}\right) \times \left(- \frac{1}{20}\right)$

Chain rule: the derivative of $e^{- \frac{x}{20}}$ keeps the whole argument, giving $e^{- \frac{x}{20}}$, then multiply by the derivative of the argument, $- \frac{1}{20}$. The constant multiplier stays in front.

$\displaystyle = \textcolor{#010106}{\frac{7 e^{- \frac{x}{20}}}{50}}$

Combine the factors into one term.

$\displaystyle \frac{d^{2}y}{dx^{2}}\;= \frac{7 \times e^{- \frac{\left(0\right)}{20}}}{50}$

Put $x = 0$ into the second derivative.

$\displaystyle = \textcolor{#010106}{\frac{7}{50}}$

Work out the exact value.

The value is positive, so the gradient is increasing as $x$ increases through that point. In words: the gradient is negative but increasing towards $0$, so the temperature falls more and more slowly: the cooling is slowing down.

Answer

(a) Rate (°C per minute): $\frac{dy}{dx} = - \frac{14}{5}$ when $x = 0$

(b) $\frac{d^{2}y}{dx^{2}} = \frac{7 e^{- \frac{x}{20}}}{50}$; Second derivative (°C per minute squared): $\frac{d^{2}y}{dx^{2}} = \frac{7}{50}$, so the gradient is negative but increasing towards $0$, so the temperature falls more and more slowly: the cooling is slowing down

$\text{depth after 3 h}$$\text{front view}$

Maths In Real Life 3

Rainwater in a barrel

When rain starts, a barrel fills quickly at first and then more slowly as the water gets deeper. A logarithm models this, and its gradient and second derivative describe the filling.

Rainwater collects in a barrel. The depth of water is $y$ cm at $x$ hours after the rain starts, where $y = 40 \ln{\left(x + 1 \right)}$, for $0 \le x \le 5$.

• (a)Find how fast the depth is rising when $x = 3$.
• (b)Find $\frac{d^{2}y}{dx^{2}}$ and its value when $x = 3$, and explain why the depth rises more and more slowly.

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Colour keyfinal result

(a)

$\displaystyle \frac{dy}{dx}\;= 40 \times \frac{1}{x + 1}$

The derivative of $\ln f(x)$ is $f'(x)$ over $f(x)$: the derivative of the argument, $1$, over the argument, $x + 1$. The constant multiplier stays in front.

$\displaystyle = \textcolor{#010106}{\frac{40}{x + 1}}$

Combine the factors into one term.

$\displaystyle \frac{dy}{dx}\;= \frac{40}{\left(3\right) + 1} = \textcolor{#010106}{10}$

Put $x = 3$ into the derivative, then work out the exact value.

(b)

$\displaystyle \frac{dy}{dx}\;= 40 \times \frac{1}{x + 1}$

The derivative of $\ln f(x)$ is $f'(x)$ over $f(x)$: the derivative of the argument, $1$, over the argument, $x + 1$. The constant multiplier stays in front.

$\displaystyle = \textcolor{#010106}{\frac{40}{x + 1}}$

Combine the factors into one term.

Differentiate the first derivative once more.

$\displaystyle \frac{d^{2}y}{dx^{2}}\;= -40\frac{1}{\left(x + 1\right)^{2}} \times 1$

Chain rule: bring the power $-1$ down, lower it by $1$ keeping the bracket, then multiply by the derivative of the bracket, $1$.

$\displaystyle = \textcolor{#010106}{- \frac{40}{\left(x + 1\right)^{2}}}$

Combine the factors into one term.

$\displaystyle \frac{d^{2}y}{dx^{2}}\;= - \frac{40}{\left(\left(3\right) + 1\right)^{2}}$

Put $x = 3$ into the second derivative.

$\displaystyle = \textcolor{#010106}{- \frac{5}{2}}$

Work out the exact value.

The value is negative, so the gradient is decreasing as $x$ increases through that point. In words: the rate at which the depth rises is decreasing: the depth rises more and more slowly.

Answer

(a) Rate (cm per hour): $\frac{dy}{dx} = 10$ when $x = 3$

(b) $\frac{d^{2}y}{dx^{2}} = - \frac{40}{\left(x + 1\right)^{2}}$; Second derivative (cm per hour squared): $\frac{d^{2}y}{dx^{2}} = - \frac{5}{2}$, so the rate at which the depth rises is decreasing: the depth rises more and more slowly

Chapter 11 gave you the rules. This chapter adds five new standard results to the list of functions you can differentiate: $\sin x$, $\cos x$, $\tan x$, $e^{x}$ and $\ln x$. Start with the three trigonometric ones. Each result below needs $x$ in radians, the unit used for every calculus statement in this chapter.

The picture shows why the results look the way they do. On the curve $y = \sin x$ the gradient at a point is the value of $\cos x$ there; on $y = \cos x$ the gradient is $-\sin x$, so the cosine curve falls where the sine is positive. The three tangents below are drawn on one window, so you can compare how steep each curve is at its marked point.

Constant multiples and sums work as before: the derivative of a sum is the sum of the derivatives, and a constant multiplier stays in front. Terms in $x^{n}$ use the power rule of Chapter 11.

When the angle is not just $x$, the angle's own rate of change multiplies the result. This is the chain rule of Chapter 11: keep the whole angle inside the outside derivative, then multiply by the derivative of the angle.

Once you have a derivative, a gradient at a given point is one substitution away. Because the angle is in radians, the angle $\dfrac{\pi}{6}$ is allowed and its sine and cosine are exact: put the angle in, use the exact trigonometric values you know, and leave the answer as a fraction or a surd. A calculator set to degrees would give the wrong gradient: set it to radians, or better, use exact values.

Two more standard results complete the toolbox. The exponential function $e^{x}$ has the remarkable property that its gradient at any point equals its height there, and the natural logarithm $\ln x$ has a gradient that gets smaller as $x$ grows. With a linear argument such as $e^{3x + 1}$ or $\ln(2x + 3)$, the chain rule brings the coefficient of $x$ into the answer.

A composite function is a function inside another function, such as $e^{x^{2}}$ (an exponential of a quadratic) or $\sin^{3} x$ (a cube of a sine). The chain rule of Chapter 11 differentiates them: differentiate the outside, keep the inside unchanged, then multiply by the derivative of the inside.