Derivatives of Trigonometric Functions

Reading

• Section 3.6

Practice problems

• Section 3.6: 5, 7, 17, 21, 27, 29, 41, 45
• To turn in (together with 3.7): 3.6 18, 28, 42

Notes

Derivatives of sine and cosine

\[\frac{d}{dx}\sin x = \cos x,\qquad \frac{d}{dx} \cos x = -\sin x\]

Watch out for the minus sign!

Proof:

• Definition of derivative: \(\displaystyle\frac{d}{dx}\sin x = \lim_{h\to 0}\frac{\sin(x+h) - \sin(x)}{h}\)
• Use trig identity: \(\sin(x+h) = \sin x \cos h + \cos x \sin h\)
• Algebra

Second derivatives:

\[\frac{d^2}{dx^2} \sin x = - \sin x, \qquad \frac{d^2}{dx^2} \cos x = - \cos x\] Both \(\sin\) and \(\cos\) satisfy the differential equation \(f''(x) = -f(x)\).

Practice problems: Derivatives of \(\sin x + \cos x\), \(x\sin(2x)\), \(\frac{x}{\cos x}\)

Derivatives of other trigonometric functions

Other trigonometric functions are written in terms of sine and cosine. We can compute their derivatives using the derivatives of \(\sin\) and \(\cos\).

\[\frac{d}{dx}\tan x = \sec^2x = 1 + \tan^2x, \qquad \frac{d}{dx}\sec x = \sec x \tan x\]

Proof:

• \(\tan x = \frac{\sin x}{\cos x}\)
• Apply quotient rule: \(\tan' x = \frac{\sin^2 x + \cos^2 x}{\cos^2 x}\)
• Numerator equal 1, so this is \(\sec^2x\).
• Alternatively, rewrite it as \(1 + \frac{\sin^2x}{\cos^2x} = 1+\tan^2x\).

Practice: Do the same for \(\sec x\).

Practice: Compute the tangent lines to \(\tan x\) at \(x=0\) and at \(x=\frac{\pi}{4}\).