Further derivatives and integrals; partial fractions

This note extends your differentiation and integration toolkit for HL. You will add derivatives of tangent, secant, cosecant, cotangent, general exponentials and logarithms and the inverse trigonometric functions. You will also learn the two integrals that give inverse trigonometric functions and how to split a rational function into partial fractions so that it can be integrated.

Key ideas

Further derivatives. Along with the chain, product and quotient rules, learn these results (angles in radians):

• \(\dfrac{d}{dx}\tan x=\sec^2x\), and \(\dfrac{d}{dx}\sec x=\sec x\tan x\).
• \(\dfrac{d}{dx}\cot x=-\operatorname{cosec}^2x\), and \(\dfrac{d}{dx}\operatorname{cosec}x=-\operatorname{cosec}x\cot x\).
• \(\dfrac{d}{dx}a^x=a^x\ln a\) for a constant \(a\gt0\), and \(\dfrac{d}{dx}\log_ax=\dfrac1{x\ln a}\).
• \(\dfrac{d}{dx}\arcsin x=\dfrac1{\sqrt{1-x^2}}\) and \(\dfrac{d}{dx}\arccos x=-\dfrac1{\sqrt{1-x^2}}\), for \(-1\lt x\lt1\).
• \(\dfrac{d}{dx}\arctan x=\dfrac1{1+x^2}\).

The chain rule extends each result. For example, \(\dfrac{d}{dx}\arctan(3x)=\dfrac{3}{1+(3x)^2}\). The pattern for the \(\log_a\) result comes from the change of base \(\log_ax=\dfrac{\ln x}{\ln a}\). The derivative of \(a^x\) comes from writing \(a^x=e^{x\ln a}\).

Where the results come from. You can derive some of these quickly, which helps you remember them. Writing \(\tan x=\dfrac{\sin x}{\cos x}\) and using the quotient rule gives \(\dfrac{\cos^2x+\sin^2x}{\cos^2x}=\dfrac1{\cos^2x}=\sec^2x\). For the inverse tangent, let \(y=\arctan x\), so that \(x=\tan y\). Differentiating with respect to \(x\) gives \(1=\sec^2y\,\dfrac{dy}{dx}\), and since \(\sec^2y=1+\tan^2y=1+x^2\), we get \(\dfrac{dy}{dx}=\dfrac1{1+x^2}\).

Integrals leading to inverse trigonometric functions. Reversing the derivative results above gives, for a constant \(a\gt0\): \[\int\frac{1}{x^2+a^2}\,dx=\frac1a\arctan\frac xa+C,\qquad \int\frac{1}{\sqrt{a^2-x^2}}\,dx=\arcsin\frac xa+C\ \ (|x|\lt a).\] Look at the shape of the denominator. A sum of squares in the denominator points to arctan. A square root of a difference points to arcsin. Also, \(\int a^x\,dx=\dfrac{a^x}{\ln a}+C\).

Constants in the numerator come outside. For instance, \(\int\dfrac{6}{x^2+9}\,dx=6\cdot\dfrac13\arctan\dfrac x3+C=2\arctan\dfrac x3+C\).

When the coefficient of \(x^2\) is not 1. To integrate \(\dfrac1{9+4x^2}\), factor out the 4 so that the denominator has the shape \(x^2+a^2\): \(\dfrac1{4\left(x^2+\frac94\right)}\). Here \(a=\tfrac32\), so the integral is \(\dfrac14\cdot\dfrac1{3/2}\arctan\dfrac{x}{3/2}+C=\dfrac16\arctan\dfrac{2x}3+C\). You can check it by differentiating with the chain rule.

Mixed rules. These new derivatives combine with the product and quotient rules in the usual way. For example, \(\dfrac{d}{dx}\big(x\arctan x\big)=\arctan x+\dfrac{x}{1+x^2}\).

Partial fractions. A rational function \(\dfrac{P(x)}{Q(x)}\) with a factorised denominator can often be written as a sum of simpler fractions. For distinct linear factors: \[\frac{px+q}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}.\] Multiply both sides by \((x-a)(x-b)\) to get \(px+q=A(x-b)+B(x-a)\). Find \(A\) and \(B\) by substituting \(x=a\) and \(x=b\), which each remove one unknown. You can also compare coefficients. If the numerator has degree equal to or greater than the denominator, divide first so that the fraction part has a smaller degree on top.

Once split, each part integrates as a logarithm: \(\int\dfrac{A}{x-a}\,dx=A\ln|x-a|+C\). Partial fractions are the key step in many integrals that look impossible at first.

An improper fraction. To integrate \(\dfrac{x^2}{x^2-1}\), the top and bottom have the same degree. Write \(\dfrac{x^2}{x^2-1}=1+\dfrac1{x^2-1}\). Then split the remaining fraction: \(\dfrac1{(x-1)(x+1)}=\dfrac12\left(\dfrac1{x-1}-\dfrac1{x+1}\right)\). So the integral is \(x+\tfrac12\ln|x-1|-\tfrac12\ln|x+1|+C\).

Choosing a method for integrals. If the denominator factorises into different linear factors, use partial fractions. If it is a sum of two squares or the square root of a difference of two squares, think inverse trigonometric. If the numerator is the derivative of the denominator, the result is a logarithm. Otherwise, consider substitution.

Definitions and formulae

Worked examples

Common errors

• Common error: forgetting the chain-rule factor, for example \(\dfrac{d}{dx}\tan(2x)=\sec^2(2x)\) instead of \(2\sec^2(2x)\).
• Common error: mixing up the signs on the "co" functions. The derivatives of \(\cot x\) and \(\operatorname{cosec}x\) both carry a minus sign.
• Common error: using \(\arctan\) for \(\dfrac1{x^2-a^2}\) or for \(\dfrac1{\sqrt{x^2+a^2}}\). The formulae only apply to the exact shapes given.
• Common error: leaving out the factor \(\dfrac1a\) in \(\int\dfrac1{x^2+a^2}\,dx\).
• Common error: attempting partial fractions on an improper fraction without dividing first, or leaving logarithms without the modulus in an indefinite integral.

Exam tip: Put your GDC in radian mode before evaluating any inverse trigonometric value. Give exact values such as \(\dfrac\pi6\) when the question allows, and a 3 s.f. decimal otherwise.

Check yourself

1. Differentiate \(\tan(5x)\) and \(\arcsin(x^2)\).
   Answer

   \(\frac{d}{dx}\tan5x=5\sec^2(5x)\). \(\frac{d}{dx}\arcsin(x^2)=\frac{2x}{\sqrt{1-x^4}}\).
2. Find \(\displaystyle\int\frac{1}{x^2+16}\,dx\) and \(\displaystyle\int_0^{3/2}\frac{1}{\sqrt{9-x^2}}\,dx\).
   Answer

   First: \(\frac14\arctan\frac x4+C\). Second: \(\left[\arcsin\frac x3\right]_0^{3/2}=\arcsin\frac12=\frac\pi6\).
3. Write \(\dfrac3{x(x+3)}\) in partial fractions and evaluate \(\displaystyle\int_1^3\frac3{x(x+3)}\,dx\).
   Answer

   \(\frac3{x(x+3)}=\frac1x-\frac1{x+3}\). Then \(\left[\ln x-\ln(x+3)\right]_1^3=(\ln3-\ln6)-(\ln1-\ln4)=\ln4-\ln2=\ln2\approx0.693\).
4. Find the gradient of \(y=3^x\) at \(x=2\).
   Answer

   \(\frac{dy}{dx}=3^x\ln3\). At \(x=2\): \(9\ln3=9.89\) (3 s.f.).