- Trigonometric Differentiation

The table below shows a list of some of the

derivatives of trigonometric functions.

Function

1st Derivative

\(\sin \) \(\cos x\)
\(\sin(u(x))\) \(u' \cos(u(x))\)
\(\cos x\) \(-\sin x\)
\(\cos (u(x))\) \(-u' \sin(u(x))\)
\(\tan x\)
\(\sec^2 x\)
\(\tan(u(x))\)
\(u' \sec^2 x\)
\(\csc x\)
\(-\csc x \cot x\)
\(\csc (u(x))\)
\(u' \csc (u(x)) \cot (u(x))\)
\(\sec x\)
\(u' \sec x \tan x\)
\(\sec (u(x))\) \(u' \sec (u(x)) \tan (u(x))\)
\(\cot x\) \(\csc^2 x\)
\(\cot (u(x))\) \(-u' \csc^2 (u(x))\)


LESSON 1

Differentiate the following w.r.t \(x\)

(a) \(y=\sec ⁡4x\)
(b) \(y=\cot ⁡5x\)
(c) \(y=3 \csc⁡(1-2x^3)\)
(d) \(\displaystyle y={1\over 2+\csc⁡(-4x)}\)
(e) \(\displaystyle y={x^2\over \sec⁡ x^3}\)
SOLUTION
(a) \(y=\sec ⁡4x\)
\(u(x)=4x → u'(x)=4\)
\(y'=4 \sec ⁡4x \tan ⁡4x\)

(b) \(y=\cot ⁡5x\)
\(u(x)=5x → u'(x)=5\)
\(y'=-5 \csc^2 ⁡5x\)

(c) \(y=3 \csc ⁡(1-2x^3)\)
\(u(x)=(1-2x^3) → u' (x)=-6x^2\)
\(y'=3[(-6x^2) \csc⁡(1-2x^3 ) \cot ⁡(1-2x^3)\)
\(=-18x^2 \csc⁡(1-2x^3) \cot⁡(1-2x^3)\)

(d) \(\displaystyle y={1\over 2+\csc⁡ (-4x)}=(2+\csc⁡ (-4x))^{-1}\)
\(u(x)=-4x → u'(x)=-4\)
\(y'=-1(2+\csc⁡ (-4x))^{-2}(4 \csc ⁡(-4x) \cot⁡(-4x))\)
\(\displaystyle =-{4 \csc ⁡(-4x) \cot ⁡(-4x)\over (2+\csc⁡(-4x))^2}\)

(e) \(\displaystyle y={x^2\over \sec⁡ x^3}\)
For \(\sec⁡ x^3: u(x)=x^3 → u'(x)=3x^2\)
Use the QUOTIENT RULE
\(\displaystyle y'={\sec⁡ x^3(2x)-(3x^2 \sec⁡ x^3 \tan⁡ x^3)(x^2)\over (sec⁡ x^3)^2}\)
\(\displaystyle ={\sec⁡ x^3 (2x-3x^4 \tan⁡ x^3)\over (sec⁡ x^3)^2}\)
\(\displaystyle ={2x-3x^4 \tan⁡ x^3\over \sec⁡ x^3}\) ⁡

LESSON 2
Given that \(y=x \tan ⁡x\), show that \(\displaystyle x^2 {d^2y\over dx^2}≡2(x^2+y^2)(1+y)\)
SOLUTION
\(\displaystyle {dy\over dx}=(1) \tan ⁡x+x \sec^2⁡x\)
  • Use the PRODUCT RULE for \(x \sec^2⁡x\)
\(\displaystyle {d^2y\over dx^2}=\sec^2⁡x+(1) \sec^2⁡x+x(2 \sec ⁡x)(\sec ⁡x \tan ⁡x)\)
\(=2 \sec^2⁡x+(2 \sec^2⁡x )x \tan ⁡x\)
\(=2 \sec^2⁡x (1+x \tan ⁡x)\)
RECALL: \(\sec^2⁡x=1+\tan^2⁡x\)
\(\displaystyle {d^2y\over dx^2}=2(1+\tan^2⁡x)(1+x \tan ⁡x)\)
\(\displaystyle x^2 {d^2 y\over dx^2}=2(x^2+x^2 \tan^2⁡x )(1+x \tan ⁡x)\)
\(=2(x^2+(x \tan ⁡x)^2)(1+x \tan ⁡x)\)
\(=2(x^2+y^2)(1+y)\)