5.15 Derivatives of further functions

The derivatives of $\sin x$, $\cos x$, $e^x$ and $\ln x$, together with the chain, product and quotient rules, are enough to differentiate a large family of functions. The remaining standard functions, the other trigonometric functions, exponentials and logarithms with any base, and the inverse trigonometric functions, complete the list of standard derivatives.

Their real importance shows when the results are read backwards. The derivative of $\arctan x$ is $\frac{1}{1 + x^2}$, a simple rational function with no trigonometry in it, so integrating $\frac{1}{1 + x^2}$

Derivatives of the trigonometric functions

The derivative of tan x

Writing $\tan x = \frac{\sin x}{\cos x}$ and using the quotient rule gives

The identity $\cos^2 x + \sin^2 x = 1$ simplifies the numerator. Since $\sec^2 x \ge 1$

The reciprocal functions

The reciprocal trigonometric functions are differentiated by writing them as negative powers and using the chain rule. For $\sec x = (\cos x)^{-1}$:

In the same way, $\operatorname{cosec} x = (\sin x)^{-1}$ gives

For $\cot x = \frac{\cos x}{\sin x}$, the quotient rule gives

These results hold only with $x$ in radians, as do those for $\sin x$ and $\cos x$. On many calculators and in some books $\operatorname{cosec} x$ is written $\csc x$.

Using the results with the chain, product and quotient rules

Second derivatives and stationary points

The new derivatives combine with everything in 5.7 and 5.8. A second derivative often simplifies with a Pythagorean identity, and stationary points are found as before by solving $f'(x) = 0$.

Exponential functions with any base

The derivative of a to the power x

The function $e^x$ is its own derivative. Any other exponential function $a^x$, with $a > 0$, can be rewritten with base $e$ using $a = e^{\ln a}$

Here $\ln a$ is a constant, so the chain rule gives

The derivative of $a^x$ is proportional to $a^x$ itself, with constant of proportionality $\ln a$. This is the defining property of exponential change: the rate of growth is proportional to the current size. When $a = e$

A population model shows the meaning of the derivative. A town of $1500$ people growing by $4\%$ a year has population $P = 1500 \times 1.04^t$ after $t$ years, so

After $10$ years the population is $1500 \times 1.04^{10} \approx 2220$ and it is growing at about $2220 \times 0.0392 \approx 87$ people per year. The growth rate is $\ln 1.04 \approx 3.92\%$

Logarithms with any base

The derivative of log base a of x

The change of base formula from 1.7 writes any logarithm in terms of natural logarithms:

Since $\frac{1}{\ln a}$ is a constant,

For example, $\frac{d}{dx}(\log x) = \frac{1}{x\ln 10}$

Logarithmic differentiation

A function such as $x^x$ has the variable in both the base and the exponent, so neither the power rule nor the rule for $a^x$ applies. Logarithmic differentiation takes the natural logarithm of both sides, which turns the exponent into a factor, and then uses implicit differentiation from 5.14.

The derivative of $x^x$ is zero when $1 + \ln x = 0$, that is at $x = e^{-1}$

Derivatives of the inverse trigonometric functions

The derivative of arcsin x

The inverse trigonometric functions are defined in 3.9 on restricted domains: $y = \arcsin x$ means $\sin y = x$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$

To write this in terms of $x$, use $\cos^2 y = 1 - \sin^2 y = 1 - x^2$

A right-angled triangle gives the same result: if $\sin y = x$, label the hypotenuse $1$ and the opposite side $x$; then the adjacent side is $\sqrt{1 - x^2}$

The derivative is undefined at $x = \pm 1$, where the denominator is $0$. The graph of $y = \arcsin x$ has vertical tangents at its end points $\left(\pm 1, \pm\frac{\pi}{2}\right)$

The derivatives of arccos x and arctan x

For $y = \arccos x$, $\cos y = x$ with $0 \le y \le \pi$. Differentiating implicitly, $-\sin y\frac{dy}{dx} = 1$

The two derivatives differ only in sign. This is expected, since $\arcsin x + \arccos x = \frac{\pi}{2}$ for every $x$ in $[-1, 1]$

For $y = \arctan x$, $\tan y = x$ with $-\frac{\pi}{2} < y < \frac{\pi}{2}$

Differentiating composites of inverse trigonometric functions

The result in (b) is a pattern worth recognising: $\frac{d}{dx}\arctan\left(\frac{x}{a}\right) = \frac{a}{a^2 + x^2}$

A rotating camera gives a practical use of the derivative of $\arctan$. A camera $20$ m from a straight road turns to follow a car travelling at $25\text{ m s}^{-1}$. If $x$ is the car's distance along the road from the point nearest the camera, the camera's angle is $\theta = \arctan\frac{x}{20}$

The camera turns fastest, at $1.25\text{ rad s}^{-1}$, as the car passes ($x = 0$), and only $0.625\text{ rad s}^{-1}$ when the car is $20$

Integrals from the new derivatives

Standard integrals

Every one of these derivatives gives a standard integral when read backwards.

Each result in the table can be checked by differentiating the right-hand side.

The formula booklet gives the last two in a more general form, with a constant $a > 0$:

These can be verified by differentiation. For the first, the chain rule gives $\frac{1}{a} \times \frac{1}{1 + \frac{x^2}{a^2}} \times \frac{1}{a} = \frac{1}{a^2 + x^2}$

There is no separate integral for $-\frac{1}{\sqrt{1 - x^2}}$

Composites with a linear function

If $\int f(x)\,dx = F(x) + C$, then for constants $a \ne 0$ and $b$

because differentiating $\frac{1}{a}F(ax + b)$ with the chain rule gives back $f(ax + b)$. This is integration by inspection (5.10) applied to the new standard integrals.

When the coefficient of $x^2$ is not $1$, take it out as a factor first so that the integrand matches the booklet form exactly.

Definite integrals

Definite integrals of these functions often give exact answers in terms of $\pi$, because the inverse trigonometric functions take values such as $\frac{\pi}{4}$ and $\frac{\pi}{6}$

Part (a) means that the area under $y = \frac{1}{1 + x^2}$ from $x = 0$ to $x = 1$

Rearranging the integrand

Many integrands do not match a standard form until they are rearranged. Three rearrangements are common: completing the square, splitting the numerator, and partial fractions. Trigonometric identities give a fourth.

Completing the square

A quadratic denominator with no real roots can be written as $(x + p)^2 + q$ with $q > 0$ by completing the square. The integrand is then a composite of $\frac{1}{a^2 + u^2}$

Splitting the numerator

When the numerator contains $x$, split the fraction so that one part has the derivative of the denominator on top (giving a logarithm) and the other part has a constant on top (giving an arctan).

Using trigonometric identities

The Pythagorean identities from 3.9, $1 + \tan^2 x = \sec^2 x$ and $1 + \cot^2 x = \operatorname{cosec}^2 x$

Partial fractions

A rational function whose denominator factorises into two distinct linear factors can be split into partial fractions, as in 1.11. Each fraction integrates to a natural logarithm.

The two denominators $x^2 - 1$ and $x^2 + 1$ look alike but lead to completely different integrals. The first factorises into real linear factors and gives logarithms; the second does not factorise and gives an arctan.

Exam-style questions on further derivatives and integrals

Self-check

• To do:

  I can derive the derivatives of $\tan x$, $\sec x$, $\operatorname{cosec} x$ and $\cot x$ from those of $\sin x$ and $\cos x$