Question

Differentiate h(t)=2cot^2(pi*t + 2)

Answer 1

chain rule

Answer 2

think cot(pit + 2) = x in ur head,
then it looks like x^2 right ?

Answer 3

yes

Answer 4

wats derivative of x^2 ?

Answer 5

1

Answer 6

nope.

use below formula :-

\(\large \frac{d}{dx}(x^n) = nx^{n-1}\)

Answer 7

:/

Answer 8

the derivative of x is 1
x^2 is
2x

Answer 9

\(\large \frac{d}{dx}(x^2) = 2x^{2-1} \)

Answer 10

You multiply x by the power as ganeshie has shown and take the power down by 1

Answer 11

ok i understand

Answer 12

good, so we have this :-

\(h(t)=2 \cot^2(pi*t + 2) \)

*think \(x = \cot(pi*t + 2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2x \times \frac{d}{dt}(x)\)

Answer 13

replace x now

Answer 14

good, so we have this :-

\(h(t)=2 \cot^2(pi*t + 2) \)

*think \(x = \cot(pi*t + 2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2x \times \frac{d}{dt}(x)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times \frac{d}{dt}(\cot(pi*t+2))\)

Answer 15

you wid me still ? :o

Answer 16

yes

Answer 17

good, keep taking the derivatives. its a chain !

Answer 18

good, so we have this :-

\(h(t)=2 \cot^2(pi*t + 2) \)

*think \(x = \cot(pi*t + 2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2x \times \frac{d}{dt}(x)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times \color{red}{\frac{d}{dt}(\cot(pi*t+2))}\)

Answer 19

that red part u still need to work,

Answer 20

*think in ur head, pi*t + 2 = x
then, it looks like cot(x) right ?

Answer 21

yea

Answer 22

wats derivative of cot(x) ?

Answer 23

derivative of cot is -cosec^2,
use it there

Answer 24

csc ^2 x

Answer 25

good, so we have this :-

\(h(t)=2 \cot^2(pi*t + 2) \)

*think \(x = \cot(pi*t + 2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2x \times \frac{d}{dt}(x)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times \color{red}{\frac{d}{dt}(\cot(pi*t+2))}\)

*think \(x = pi*t+2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times (-\csc^2x )\color{red}{\frac{d}{dt}(x)}\)

Answer 26

good, so we have this :-

\(h(t)=2 \cot^2(pi*t + 2) \)

*think \(x = \cot(pi*t + 2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2x \times \frac{d}{dt}(x)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times \color{red}{\frac{d}{dt}(\cot(pi*t+2))}\)

*think \(x = pi*t+2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times (-\csc^2x )\color{red}{\frac{d}{dt}(x)}\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times (-\csc^2(pi*t+2) )\color{red}{\frac{d}{dt}(pi*t+2)}\)

Answer 27

one last round of derivative,
then we're done.

Answer 28

wats derivative of \(pi*t + 2\) ?

Answer 29

idk

Answer 30

its just \(pi\)

Answer 31

good, so we have this :-

\(h(t)=2 \cot^2(pi*t + 2) \)

*think \(x = \cot(pi*t + 2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2x \times \frac{d}{dt}(x)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times \color{red}{\frac{d}{dt}(\cot(pi*t+2))}\)

*think \(x = pi*t+2)\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times (-\csc^2x )\color{red}{\frac{d}{dt}(x)}\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times (-\csc^2(pi*t+2) )\color{red}{\frac{d}{dt}(pi*t+2)}\)

\(\large \frac{d}{dt}(h(t))=2 \times 2\cot(pi*t+2) \times (-\csc^2(pi*t+2) \times pi \)

Answer 32

see if that makes some sense

Answer 33

yes

Answer 34

good, remember this :-

\(\large \frac{d}{d\color{red}{x}}(cx) = c\)

\(\large \frac{d}{d\color{red}{t}}(ct) = c\)

Answer 35

we're done wid the differentiation btw.

Answer 36

yay thank you so much!