ma 241 si: sections 3.5 and 3.6 - derivatives of trig functions and the chain rule\nname: \nsection #\nfor…
ma 241 si: sections 3.5 and 3.6 - derivatives of trig functions and the chain rule\nname: \nsection #\nfor this assignment to count as extra credit, it must be submitted online with the si leaders signature at the bottom. also, before leaving the si session, be sure to ask your si leader for clarification on any topics covered thus far in your class and any questions you have pertaining to homework problems.\n1. complete the chart.\nderivatives of trig functions\n$\frac{d}{dx}(sin x)=$ $\frac{d}{dx}(cos x)=$\n$\frac{d}{dx}(\tan x)=$ $\frac{d}{dx}(cot x)=$\n$\frac{d}{dx}(sec x)=$ $\frac{d}{dx}(csc x)=$\nalso keep in mind the relationships between the trig functions as well as the trigonometric values of special angles. fill in the blanks.\n$\frac{sin x}{cos x}=$ $\frac{cos x}{sin x}=$\n$\frac{1}{sec x}=$ $\frac{1}{csc x}=$\n$\frac{1}{cot x}=$ $\frac{1}{\tan x}=$\n2. find $\frac{dy}{dx}$ and simplify: $y = (sin x+cos x)csc x$\n3. find $\frac{dy}{dx}$ and simplify: $y=\frac{5}{sin(x)}-\frac{2}{cos(x)}$
Answer
Explanation:
Step1: Recall derivative formulas
The derivative of $\sin x$ is $\cos x$, i.e., $\frac{d}{dx}(\sin x)=\cos x$. The derivative of $\cos x$ is $-\sin x$, i.e., $\frac{d}{dx}(\cos x)=-\sin x$. The derivative of $\tan x=\sec^{2}x$, i.e., $\frac{d}{dx}(\tan x)=\sec^{2}x$. The derivative of $\cot x = -\csc^{2}x$, i.e., $\frac{d}{dx}(\cot x)=-\csc^{2}x$. The derivative of $\sec x=\sec x\tan x$, i.e., $\frac{d}{dx}(\sec x)=\sec x\tan x$. The derivative of $\csc x=-\csc x\cot x$, i.e., $\frac{d}{dx}(\csc x)=-\csc x\cot x$.
Step2: Recall trig - identities
$\frac{\sin x}{\cos x}=\tan x$, $\frac{\cos x}{\sin x}=\cot x$, $\frac{1}{\sec x}=\cos x$, $\frac{1}{\csc x}=\sin x$, $\frac{1}{\cot x}=\tan x$, $\frac{1}{\tan x}=\cot x$. For the unit - circle, at $x = \frac{\pi}{2}$, the point is $(0,1)$; at $x=\frac{\pi}{3}$, the point is $(\frac{1}{2},\frac{\sqrt{3}}{2})$; at $x = \frac{\pi}{4}$, the point is $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$; at $x=\frac{\pi}{6}$, the point is $(\frac{\sqrt{3}}{2},\frac{1}{2})$.
Step3: Solve problem 2
First, expand $y = (\sin x+\cos x)\csc x=\sin x\csc x+\cos x\csc x$. Since $\sin x\csc x = 1$ and $\cos x\csc x=\frac{\cos x}{\sin x}=\cot x$, then $y = 1+\cot x$. The derivative $\frac{dy}{dx}=\frac{d}{dx}(1)+\frac{d}{dx}(\cot x)$. Since $\frac{d}{dx}(1) = 0$ and $\frac{d}{dx}(\cot x)=-\csc^{2}x$, so $\frac{dy}{dx}=-\csc^{2}x$.
Step4: Solve problem 3
Rewrite $y=\frac{5}{\sin x}-\frac{2}{\cos x}=5\csc x - 2\sec x$. Using the sum - rule of differentiation $\frac{dy}{dx}=\frac{d}{dx}(5\csc x)-\frac{d}{dx}(2\sec x)$. Since $\frac{d}{dx}(a\cdot f(x))=a\frac{d}{dx}(f(x))$ for a constant $a$, and $\frac{d}{dx}(\csc x)=-\csc x\cot x$, $\frac{d}{dx}(\sec x)=\sec x\tan x$. Then $\frac{dy}{dx}=5(-\csc x\cot x)-2(\sec x\tan x)=- 5\csc x\cot x-2\sec x\tan x$.
Answer:
| Derivatives of Trig Functions | |
|---|---|
| $\frac{d}{dx}(\sin x)=\cos x$ | $\frac{d}{dx}(\cos x)=-\sin x$ |
| $\frac{d}{dx}(\tan x)=\sec^{2}x$ | $\frac{d}{dx}(\cot x)=-\csc^{2}x$ |
| $\frac{d}{dx}(\sec x)=\sec x\tan x$ | $\frac{d}{dx}(\csc x)=-\csc x\cot x$ |
Unit - circle points: At $x=\frac{\pi}{2}:(0,1)$; at $x = \frac{\pi}{3}:(\frac{1}{2},\frac{\sqrt{3}}{2})$; at $x=\frac{\pi}{4}:(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$; at $x=\frac{\pi}{6}:(\frac{\sqrt{3}}{2},\frac{1}{2})$
Trig identities: $\frac{\sin x}{\cos x}=\tan x$, $\frac{\cos x}{\sin x}=\cot x$, $\frac{1}{\sec x}=\cos x$, $\frac{1}{\csc x}=\sin x$, $\frac{1}{\cot x}=\tan x$, $\frac{1}{\tan x}=\cot x$
Problem 2: $\frac{dy}{dx}=-\csc^{2}x$
Problem 3: $\frac{dy}{dx}=-5\csc x\cot x - 2\sec x\tan x$