Trigo 2:Chain Rule

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Mathematics

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12th Grade

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Thavarajah Selvarajah

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14 Slides • 40 Questions

Trigo 2:

Chain Rule

by Thavarajah Selvarajah

For the following, decide whether

chain rule ( PIT-STOP 2) is needed?

When

chain rule ( PIT-STOP 2) is needed?

Multiple Choice

Given $y=\sin\left(1-3x\right)$ ,then $\frac{d}{dx}\left[\sin\left(1-3x\right)\right]$ =

Yes

Multiple Choice

Given $y=\sin\left(3x\right)$ ,then $\frac{d}{dx}\left(\sin3x\right)$ =

Yes

Multiple Choice

Given $y=3\sin\left(x\right)$ ,then $\frac{d}{dx}\left(3\sin x\right)$ =

Yes

Multiple Choice

Given $y=3\sin\frac{2}{3}x$ ,then $\frac{d}{dx}\left(3\sin\frac{2}{3}x\right)$ =

Yes

Multiple Choice

Given $y=3\sin w$ ,then $\frac{d}{dw}\left(3\sin w\right)$ =

Yes

WENT WELL?

COULD YOU DISTINGUISH WHICH ONES NEEDS PIT-STOP 2

( CHAIN RULE) ?

LEVEL 2 : THE PROCESS of Chain Rule :) 

Multiple Choice

What is $p?$

$\sin\ \left(1-3x\right)$

$-3x$

$1-3x$

$3x$

Multiple Choice

What is $q?$

$\sin\ u$

$\cos\ u$

$\sin\ x$

$\cos\ x$

Multiple Choice

What is $r?$

$\cos\ \left(1-3x\right)$

$-3\cos\ u$

$-3\cos\ \left(1-3x\right)$

$\cos\ x$

Multiple Choice

Given is 1st step of Chain Rule. What is $a?$

$u$

$\tan\ 2x$

$2x$

Multiple Choice

Given is continuation step of previous Chain Rule. What is $b$ and $c?$

$b=2,\ c=\tan\ u$

$b=-2,\ c=\sec^2u$

$b=2,\ c=\sec^2u$

Multiple Choice

Given is first few steps of Chain Rule differentiation of the function. What is $p?$

$\frac{2}{3}\operatorname{cosec}\ u$

$\operatorname{cosec}\ u$

$\frac{2}{3}\operatorname{cosec}\ \left(1-2x\right)$

Multiple Choice

Given is the continuation steps of previous Chain Rule differentiation of the function. What is $q?$

$-\frac{4}{3}\operatorname{cosec}u\ \cot\ u$

$-\frac{4}{3}\operatorname{cosec}\left(1-2x\right)\cot\ \left(1-2x\right)$

$\frac{2}{3}\operatorname{cosec}\ \left(1-2x\right)$

$\frac{4}{3}\operatorname{cosec}\left(1-2x\right)\cot\ \left(1-2x\right)$

ARE YOU CONFIDENT DOING THE QUESTIONS INDEPENDENTLY?

LEVEL 3

DOING PIT-STOP 2 (CHAIN RULE)

INDEPENDENTLY

ARE YOU SURE

YOU ARE READY?

LET'S DO THIS CHAMPS......

Multiple Choice

Given $y=\sin5x$ .Find $\frac{\text{d}y}{\text{d}x}$

$\cos5x$

$-\cos5x$

$5\cos5x$

$-5\cos5x$

Multiple Choice

Given $y=\frac{1}{5}\sin5x$ .Find $\frac{\text{d}y}{\text{d}x}$

$\cos5x$

$\frac{1}{5}\cos5x$

$-\frac{1}{5}\cos5x$

$-\cos5x$

Multiple Choice

Given $y=\cos2x$ .Find $\frac{\text{d}y}{\text{d}x}$

$\sin2x$

$-\sin2x$

$2\sin2x$

$-2\sin2x$

Multiple Choice

Given $y=\cot6x$ .Find $\frac{\text{d}y}{\text{d}x}$

$\operatorname{cosec}^26x$

$-\operatorname{cosec}^26x$

$-6\operatorname{cosec}^26x$

$6\operatorname{cosec}^26x$

Multiple Choice

Given $y=\frac{1}{6}\cot6x$ .Find $\frac{\text{d}y}{\text{d}x}$

$\operatorname{cosec}^26x$

$-\operatorname{cosec}^26x$

$-6\operatorname{cosec}^26x$

$6\operatorname{cosec}^26x$

Multiple Choice

Given $y=\frac{1}{6}\tan\left(1-6x\right)$ .Find $\frac{\text{d}y}{\text{d}x}$

$\sec^2\left(1-6x\right)$

$-\sec^2\left(1-6x\right)$

$\frac{1}{6}\sec^2\left(1-6x\right)$

$-\frac{1}{6}\sec^2\left(1-6x\right)$

Multiple Choice

Given $y=-\frac{1}{2}\cos2x$ .Find $\frac{\text{d}y}{\text{d}x}$

$\sin2x$

$-\frac{1}{2}\sin2x$

$\frac{1}{2}\sin2x$

$-\sin2x$

Multiple Choice

Given $y=\frac{1}{5}\sin\left(1-5x\right)$ .Find $\frac{\text{d}y}{\text{d}x}$

$\frac{1}{5}\cos\left(1-5x\right)$

$-\frac{1}{5}\cos\left(1-5x\right)$

$-\cos\left(1-5x\right)$

$\cos\left(1-5x\right)$

Multiple Choice

Given $y=\frac{2}{5}\sin\left(5x-1\right)$ .Find $\frac{\text{d}y}{\text{d}x}$

$2\cos\left(5x-1\right)$

$\frac{2}{5}\cos\left(5x-1\right)$

$-\frac{2}{5}\cos\left(5x-1\right)$

$-2\cos\left(5x-1\right)$

Multiple Choice

Given $y=\frac{1}{2}\cos\left(1-2x\right)$ , find $\frac{\text{d}y}{\text{d}x}$

$\frac{1}{2}\sin\left(1-2x\right)$

$-\frac{1}{2}\sin\left(1-2x\right)$

$\sin\left(1-2x\right)$

$-\sin\left(1-2x\right)$

Multiple Choice

Given $y=-\frac{1}{2}\cos\left(2x-1\right)$ , find $\frac{\text{d}y}{\text{d}x}$

$\frac{1}{2}\sin\left(2x-1\right)$

$-\frac{1}{2}\sin\left(2x-1\right)$

$\sin\left(2x-1\right)$

$-\sin\left(2x-1\right)$

Multiple Choice

Given $y=-\frac{1}{6}\cot6x$ , find $\frac{\text{d}y}{\text{d}x}$

$\operatorname{cosec}^26x$

$-\frac{1}{6}\operatorname{cosec}^26x$

$\frac{1}{6}\operatorname{cosec}^26x$

$-\operatorname{cosec}^26x$

Multiple Choice

Given $y=\frac{1}{6}\cot\left(1-6x\right)$

$\frac{1}{6}\operatorname{cosec}^2\left(1-6x\right)$

$-\operatorname{cosec}^2\left(1-6x\right)$

$\operatorname{cosec}^2\left(1-6x\right)$

$-\frac{1}{6}\operatorname{cosec}^2\left(1-6x\right)$

Multiple Choice

Given $y=\sec7x$ ,find $\frac{\text{d}y}{\text{d}x}$

$\sec7x\tan7x$

$7\sec7x\tan7x$

$-7\sec7x\tan7x$

$\frac{1}{7}\sec7x\tan7x$

TIRED??

Multiple Choice

Given $y=\frac{1}{7}\sec7x$

$\sec7x\tan7x$

$7\sec7x\tan7x$

$-\sec7x\tan7x$

$-7\sec7x\tan7x$

Multiple Choice

Given $y=-\frac{1}{7}\sec7x$

$-\sec7x\tan7x$

$\sec7x\tan7x$

$\frac{1}{7}\sec7x\tan7x$

$-7\sec7x\tan7x$

Multiple Choice

Given $y=\frac{1}{7}\sec\left(1-7x\right)$ ,find $\frac{dy}{dx}$

$\sec\left(1-7x\right)\tan\left(1-7x\right)$

$-\sec\left(1-7x\right)\tan\left(1-7x\right)$

$\frac{1}{7}\sec\left(1-7x\right)\tan\left(1-7x\right)$

$-\frac{1}{7}\sec\left(1-7x\right)\tan\left(1-7x\right)$

Multiple Choice

Given $y=-\frac{1}{7}\sec\left(1-7x\right)$ ,find $\frac{dy}{dx}$

$\sec\left(1-7x\right)\tan\left(1-7x\right)$

$-\sec\left(1-7x\right)\tan\left(1-7x\right)$

$\frac{1}{7}\sec\left(1-7x\right)\tan\left(1-7x\right)$

$-\frac{1}{7}\sec\left(1-7x\right)\tan\left(1-7x\right)$

Multiple Choice

Given $y=\operatorname{cosec}\ 8x$ , find $\frac{dy}{dx}$

$-\operatorname{cosec}8x\cot8x$

$8\operatorname{cosec}8x\cot8x$

$-8\operatorname{cosec}8x\cot8x$

$\operatorname{cosec}8x\cot8x$

Multiple Choice

Given $y=\frac{1}{8}\operatorname{cosec}\ 8x$ , find $\frac{dy}{dx}$

$-\operatorname{cosec}8x\cot8x$

$8\operatorname{cosec}8x\cot8x$

$-8\operatorname{cosec}8x\cot8x$

$\operatorname{cosec}8x\cot8x$

Multiple Choice

Given $y=-\frac{1}{8}\operatorname{cosec}\ 8x$ , find $\frac{dy}{dx}$

$-\operatorname{cosec}8x\cot8x$

$8\operatorname{cosec}8x\cot8x$

$-8\operatorname{cosec}8x\cot8x$

$\operatorname{cosec}8x\cot8x$

Multiple Choice

Given $y=-\frac{1}{8}\operatorname{cosec}\ \left(8x-1\right)$ , find $\frac{dy}{dx}$

$-\operatorname{cosec}\left(8x-1\right)\cot\left(8x-1\right)$

$8\operatorname{cosec}\left(8x-1\right)\cot\left(8x-1\right)$

$-8\operatorname{cosec}\left(8x-1\right)\cot\left(8x-1\right)$

$\operatorname{cosec}\left(8x-1\right)\cot\left(8x-1\right)$

Multiple Choice

Given $y=\frac{1}{8}\operatorname{cosec}\ \left(1-8x\right)$ , find $\frac{dy}{dx}$

$-\operatorname{cosec}\left(1-8x\right)\cot\left(1-8x\right)$

$8\operatorname{cosec}\left(1-8x\right)\cot\left(1-8x\right)$

$-8\operatorname{cosec}\left(1-8x\right)\cot\left(1-8x\right)$

$\operatorname{cosec}\left(1-8x\right)\cot\left(1-8x\right)$

Multiple Choice

Given $y=\tan3x$ ,find $\frac{dy}{dx}$

$\sec^23x$

$3\sec^23x$

$\sec^2x$

$3\sec^2x$

Multiple Choice

Given $y=\frac{1}{3}\tan3x$ ,find $\frac{dy}{dx}$

$\sec^23x$

$3\sec^23x$

$\sec^2x$

$3\sec^2x$

Multiple Choice

Given $y=-\frac{1}{3}\tan3x$ ,find $\frac{dy}{dx}$

$\sec^23x$

$3\sec^23x$

$-\sec^23x$

$-3\sec^23x$

Multiple Choice

Given $y=-\frac{1}{3}\tan\left(1-3x\right)$ ,find $\frac{dy}{dx}$

$\sec^2\left(1-3x\right)$

$3\sec^2\left(1-3x\right)$

$-\frac{1}{3}\sec^2\left(1-3x\right)$

$-3\sec^2\left(1-3x\right)$

Multiple Choice

Given $y=-\frac{1}{3}\tan\left(3x-1\right)$ ,find $\frac{dy}{dx}$

$\sec^2\left(1-3x\right)$

$-\sec^2\left(1-3x\right)$

$-\frac{1}{3}\sec^2\left(1-3x\right)$

$-3\sec^2\left(1-3x\right)$

WHAT YOU SHOULD HAVE UNLOCKED HERE?

1. able to identify which needs PIT-STOP 2 (Chain rule)

2. How to carry out the process independently

3. Please review the questions that you are unsure before trying the next subsection

Trigo 2:

Chain Rule

by Thavarajah Selvarajah

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