\[\frac{d}{dx}\sin x = \cos x,\qquad \frac{d}{dx} \cos x = -\sin x\]
Watch out for the minus sign!
Proof:
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 \(\sin x + \cos x\), \(x\sin(2x)\), \(\frac{x}{\cos x}\)
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:
Practice: Do the same for \(\sec x\).
Practice: Compute the tangent lines to \(\tan x\) at \(x=0\) and at \(x=\frac{\pi}{4}\).