Prerequisite
- Limit
- Power Rule
- Chain Rule
- Implicit Differenciation
- Trig Identities
Key Points
- Any trignometric functions that starts with c will be negative
- cos, csc, cot (same in arc functions)
Basic Trigonometric Functions
sin
\[\begin{align} y &= \sin (x) \\ \frac{dy}{dx} &= \lim_{h \to 0}{ \frac{\sin(x+h) - \sin (x)}{h} } \\ &= \lim_{h \to 0}{ \frac{2 \sin(\frac{x + h - x}{2})\cos(\frac{x + h + x}{2})}{h} } \\ &= \lim_{h \to 0}{ \frac{2 \sin(\frac{h}{2})\cos(\frac{2x + h}{2})}{h} } \\ &= \lim_{h \to 0}{ \frac{\sin(\frac{h}{2})\cos(x + \frac{h}{2})}{h/2} } \\ &= \lim_{h \to 0}{ \frac{\sin(\frac{h}{2})}{h/2} \cos\left(x + \frac{h}{2}\right)} \\ &= \lim_{h \to 0}{ 1 \times \cos\left(x + 0\right)} \\ &= \cos(x) \end{align}\]cos
\[\begin{align} y &= \cos (x) \\ \frac{dy}{dx} &= \lim_{h \to 0}{ \frac{\cos(x+h) - \cos (x)}{h} } \\ &= \lim_{h \to 0}{ \frac{\cos(x)\cos(h)-\sin(x)\sin(h)-\cos(x)}{h} } \\ &= \lim_{h \to 0}{ \frac{\cos(x)(\cos(h)-1)-\sin(x)\sin(h)}{h} } \\ &= \lim_{h \to 0}{ \frac{\cos(x)(\cos(h)-1)}{h} - \frac{\sin(x)\sin(h)}{h} } \\ &= \lim_{h \to 0}{ \cos(x) \frac{(\cos(h)-1)( \cos(h)+1) }{ h(\cos(h)+1) } - \sin(x) \cdot \frac{ \sin(h) }{ h } } \\ &= \lim_{h \to 0}{ \cos(x) \frac{\cos^2(h)-1}{h(\cos(h)+1)} - \sin(x)\cdot\frac{\sin(h)}{h} } \\ &= \lim_{h \to 0}{ \cos(x) \frac{-\sin^2(h)}{h(\cos(h)+1)} - \sin(x)\cdot\frac{\sin(h)}{h} } \\ &= \lim_{h \to 0}{ \cos(x) \frac{\sin(h)}{h} \cdot \frac{-\sin(h)}{(\cos(h)+1)} - \sin(x) \cdot \frac{\sin(h)}{h} } \\ &= \lim_{h \to 0}{ \cos(x) \cdot 1 \cdot \frac{0}{2} - \sin(x) \cdot 1 } \\ &= \lim_{h \to 0}{ 0 - \sin(x) } \\ &= - \sin(x) \end{align}\]tan
\[\begin{align} y &= \tan (x) = \frac{\sin(x)}{\cos(x)}=\sin(x)(\cos(x))^{-1} \\ \frac{dy}{dx} &= \frac{d}{dx}\bigg[\sin(x)\bigg]\cdot(\cos(x))^{-1} + \sin(x)\cdot\frac{d}{dx}\bigg[(\cos(x))^{-1}\bigg] \\ &= \cos(x)\cdot(\cos(x))^{-1} + \sin(x)\cdot(-1)(\cos(x))^{-2}\frac{d}{dx}\bigg[\cos(x)\bigg] \\ &= 1 + \sin(x)\cdot(-1)(\cos(x))^{-2}\cdot(-\sin(x)) \\ &= 1 + \sin^2(x)\cdot(\cos(x))^{-2} \\ &= 1 + \frac{\sin^2(x)}{\cos^2(x)} \\ &= 1 + \tan^2(x) \quad \leftarrow \text{trig identity}\\ &= \sec^2(x) \end{align}\]Reciprocal Trigonometric Functions
csc
\[\begin{align} y &= \csc (x) = \frac{1}{\sin(x)}=(\sin(x))^{-1} \\ \frac{dy}{dx} &= \frac{d}{dx}\bigg[(\sin(x))^{-1}\bigg]\\ &= (-1)(\sin(x))^{-2}\frac{d}{dx}\bigg[\sin(x)\bigg] \\ &= -(\sin(x))^{-2}\cos(x) \\ &= -\frac{\cos(x)}{\sin^2(x)} \\ &= -\frac{1}{\sin(x)}\cdot\frac{\cos(x)}{\sin(x)} \\ &= -\csc(x)\cot(x) \\ \end{align}\]sec
\[\begin{align} y &= \sec (x) = \frac{1}{\cos(x)}=(\cos(x))^{-1} \\ \frac{dy}{dx} &= \frac{d}{dx}\bigg[(\cos(x))^{-1}\bigg]\\ &= (-1)(\cos(x))^{-2}\frac{d}{dx}\bigg[\cos(x)\bigg] \\ &= -(\cos(x))^{-2}\cdot(-\sin(x)) \\ &= \frac{\sin(x)}{\cos^2(x)} \\ &= \frac{1}{\cos(x)}\cdot\frac{\sin(x)}{\cos(x)} \\ &= \sec(x)\tan(x) \\ \end{align}\]cot
\[\begin{align} y &= \cot (x) = \frac{\cos(x)}{\sin(x)}=\cos(x)(\sin(x))^{-1} \\ \frac{dy}{dx} &= \frac{d}{dx}\bigg[\cos(x)\bigg]\cdot(\sin(x))^{-1} + \cos(x)\cdot\frac{d}{dx}\bigg[(\sin(x))^{-1}\bigg] \\ &= - \sin(x)\cdot(\sin(x))^{-1} + \cos(x)\cdot(-1)(\sin(x))^{-2}\frac{d}{dx}\bigg[\sin(x)\bigg] \\ &= -1 + \cos(x)\cdot(-1)(\sin(x))^{-2}\cos(x) \\ &= -1 - \cos^2(x)\cdot(\sin(x))^{-2} \\ &= -1 - \frac{\cos^2(x)}{\sin^2(x)} \\ &= -1 - \cot^2(x) \\ &= -(1 + \cot^2(x)) \quad \leftarrow \text{trig identity} \\ &= -\csc^2(x) \end{align}\]Inverse Trigonometric Functions
arcsin
\[\begin{align} y &= \sin(\theta) \\ x &= \cos(\theta) \\ x&^2+y^2=1 \\ x^2 &= 1 - y^2 \\ x &= \sqrt{1 - y^2} \\ \frac{dx}{dy} &= \frac{-\cancel{2}y}{\cancel{2}\sqrt{1-y^2}} \\ \frac{dx}{d\theta}\cdot\frac{d\theta}{dy} &= \frac{-y}{\sqrt{1-y^2}} \\ \frac{d\cos(\theta)}{d\theta}\cdot\frac{d\theta}{dy} &= \frac{-y}{\sqrt{1-y^2}} \\ \cancel{(-\sin(\theta))}\cdot\frac{d\theta}{dy} &= \frac{\cancel{-\sin(\theta)}}{\sqrt{1-y^2}} \\ \frac{d\theta}{dy} &= \frac{1}{\sqrt{1-y^2}} \end{align}\]arccos
\[\begin{align} y &= \sin(\theta) \\ x &= \cos(\theta) \\ x&^2+y^2=1 \\ y^2 &= 1 - x^2 \\ y &= \sqrt{1 - x^2} \\ \frac{dy}{dx} &= \frac{-\cancel{2}x}{\cancel{2}\sqrt{1-x^2}} \\ \frac{dy}{d\theta}\cdot\frac{d\theta}{dx} &= -\frac{x}{\sqrt{1-x^2}} \\ \frac{d\sin(\theta)}{d\theta}\cdot\frac{d\theta}{dx} &= -\frac{x}{\sqrt{1-x^2}} \\ \cancel{\cos(\theta)}\cdot\frac{d\theta}{dx} &= -\frac{\cancel{\cos(\theta)}}{\sqrt{1-x^2}} \\ \frac{d\theta}{dx} &= -\frac{1}{\sqrt{1-x^2}} \end{align}\]arctan
\[\begin{align} \theta &= \tan^{-1}(x) \\ \tan(\theta) &= x \\ \frac{d}{dx}\bigg[\tan(\theta)\bigg]&=1 \\ \sec^2(\theta)\frac{d\theta}{dx}&=1 \\ \frac{d\theta}{dx}&=\frac{1}{\sec^2(\theta)} \\ &=\frac{1}{1+\tan^2(\theta)} \\ &=\frac{1}{1+x^2} \\ \end{align}\]arccsc
\[\begin{align} \cos^2(\theta) + \sin^2(\theta) &= 1 \\ \cos^2(\theta) &= 1 - \sin^2(\theta) \\ \cos^2(\theta) &= 1 - \frac{1}{\csc^2(\theta)} \\ \cos(\theta) &= \sqrt{1 - \frac{1}{\csc^2(\theta)}} \\ \end{align}\]manipulate trig
\[\begin{align} \theta &= \csc^{-1}(x) \\ \csc(\theta) &= x \\ \frac{d}{dx}\bigg[\csc(\theta)\bigg]&=1 \\ -\csc(\theta)\cot(\theta)\frac{d\theta}{dx}&=1 \\ \frac{d\theta}{dx}&=-\frac{1}{\csc(\theta)\cot(\theta)} \\ &=-\frac{1}{\csc(\theta)\cos(\theta)\csc(\theta)} \\ &=-\frac{1}{\csc^2(\theta)\cos(\theta)} \\ &=-\frac{1}{x^2\cos(\theta)} \\ &=-\frac{1}{x^2\sqrt{1 - \frac{1}{\csc^2(\theta)}}} \\ &=-\frac{1}{x^2\sqrt{1 - \frac{1}{x^2}}} \\ &=-\frac{1}{x^2\sqrt{1 - x^{-2}}} \\ \end{align}\]arcsec
\[\begin{align} \cos^2(\theta) + \sin^2(\theta) &= 1 \\ \sin^2(\theta) &= 1 - \cos^2(\theta) \\ \sin^2(\theta) &= 1 - \frac{1}{\sec^2(\theta)} \\ \sin(\theta) &= \sqrt{1 - \frac{1}{\sec^2(\theta)}} \\ \end{align}\]manipulate trig
\[\begin{align} \theta &= \sec^{-1}(x) \\ \sec(\theta) &= x \\ \frac{d}{dx}\bigg[\sec(\theta)\bigg]&=1 \\ \sec(\theta)\tan(\theta)\frac{d\theta}{dx}&=1 \\ \frac{d\theta}{dx}&=\frac{1}{\sec(\theta)\tan(\theta)} \\ &=\frac{1}{\sec(\theta)\sin(\theta)\sec(\theta)} \\ &=\frac{1}{\sec^2(\theta)\sin(\theta)} \\ &=\frac{1}{x^2\sin(\theta)} \\ &=\frac{1}{x^2\sqrt{1 - \frac{1}{\sec^2(\theta)}}} \\ &=\frac{1}{x^2\sqrt{1 - \frac{1}{x^2}}} \\ &=\frac{1}{x^2\sqrt{1 - x^{-2}}} \\ \end{align}\]