## - 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: $$\sec^2⁡x=1+\tan^2⁡x$$
$$\displaystyle {d^2y\over dx^2}=2(1+\tan^2⁡x)(1+x \tan ⁡x)$$
$$\displaystyle x^2 {d^2 y\over dx^2}=2(x^2+x^2 \tan^2⁡x )(1+x \tan ⁡x)$$
$$=2(x^2+(x \tan ⁡x)^2)(1+x \tan ⁡x)$$
$$=2(x^2+y^2)(1+y)$$