3.5.2 Derivatives of Trigonometric Functions

Derivative of Tangent ( (\tan x) )

• Rewrite the function in a form that is easy to differentiate:
  • $\tan x = \frac{\sin x}{\cos x}$
• Apply the Quotient Rule $\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2}$ with
  • Numerator: (f = \sin x) ⇒ (f' = \cos x)
  • Denominator: (g = \cos x) ⇒ (g' = -\sin x)
• Substitute and simplify:
  • $\tan' x = \frac{\cos x\,(\cos x)\; - \;\sin x\,(-\sin x)}{(\cos x)^2}$
  • $= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$
  • $= \frac{1}{\cos^2 x}$ (using (\sin^2 x + \cos^2 x = 1))
  • $= \sec^2 x$
• Result: $\boxed{\dfrac{d}{dx}[\tan x] = \sec^2 x}$

Standard First-Order Trigonometric Derivatives (Reference)

• $\frac{d}{dx}[\sin x] = \cos x$
• $\frac{d}{dx}[\cos x] = -\sin x$
• $\frac{d}{dx}[\tan x] = \sec^2 x$ (derived above)
• $\frac{d}{dx}[\cot x] = -\csc^2 x$
• $\frac{d}{dx}[\sec x] = \sec x\tan x$
• $\frac{d}{dx}[\csc x] = -\csc x\cot x$

(The instructor notes you may re-derive each one with similar sine–cosine rewrites plus product/quotient rules.)

Worked Example 1: Derivative of (\sec x \;\csc x)

• Recognize a product ⇒ use the Product Rule $(uv)' = u'v + uv'$
• Identify components:
  • (u = \sec x) ⇒ (u' = \sec x\tan x)
  • (v = \csc x) ⇒ (v' = -\csc x\cot x)
• Assemble derivative:
  • $\frac{d}{dx}[\sec x\,\csc x] = \sec x\tan x\,(\csc x) + \sec x\,(-\csc x\cot x)$
• Rewrite every trig function in sine–cosine form to expose cancellations:
  • $\sec x = \frac{1}{\cos x}, \quad \tan x = \frac{\sin x}{\cos x}, \quad \csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x}$
  • First term: $\frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot \frac{1}{\sin x} = \frac{1}{\cos^2 x}$
  • Second term (carry out the minus sign):
    $-\left( \frac{1}{\cos x}\right) !\left( \frac{1}{\sin x}\right) !\left( \frac{\cos x}{\sin x}\right) = -\frac{1}{\sin^2 x}$
• Convert back to standard trig functions:
  • $\frac{1}{\cos^2 x} = \sec^2 x$
  • $\frac{1}{\sin^2 x} = \csc^2 x$
• Final result:
  • $\boxed{\frac{d}{dx}[\sec x\,\csc x] = \sec^2 x - \csc^2 x}$

Worked Example 2: Second Derivative of (y = \csc x)

1. First derivative (already known rule):
   • $y' = \frac{d}{dx}[\csc x] = -\csc x\,\cot x$
2. Second derivative: differentiate (-\csc x\,\cot x) via the Product Rule.
   • Let (u = -\csc x) and (v = \cot x)
   • Derivatives:
     • $u' = -(\csc x)' = -( -\csc x\cot x ) = \csc x\cot x$
     • $v' = (\cot x)' = -\csc^2 x$
   • Combine:
     $y'' = u'v + uv' = (\csc x\cot x)(\cot x) + (-\csc x)(-\csc^2 x)$
     $= \csc x\,\cot^2 x + \csc x\,\csc^2 x$
3. Factor common (\csc x):
   • $y'' = \csc x\,(\cot^2 x + \csc^2 x)$
4. Final compact form:
   • $\boxed{y'' = \csc x\left( \cot^2 x + \csc^2 x \right)}$

Key Take-Aways & Study Hints

• Memorize the six first-order trig derivatives; they appear frequently and save time during exams.
• When proofs or simplifications are needed, convert unfamiliar products/quotients into sine–cosine form to expose cancellations and Pythagorean identities.
• Product Rule and Quotient Rule are the backbone for building higher-order trig derivatives.
• Pythagorean identities (e.g.
  $\sin^2 x + \cos^2 x = 1,\; 1 + \tan^2 x = \sec^2 x,\; 1 + \cot^2 x = \csc^2 x$)
  are essential tools for simplification.
• Higher-order derivatives often factor nicely; look for common factors such as (\csc x) or (\sec x) when cleaning up results.