Question

Find the derivative of the function.
y=sin(πx)^2

Answer 1

\(y=\sin(πx)^2\)

OR

\(y=\sin^2(πx)\)

Answer 2

well, do it then!

Answer 3

Jk the first one my b

Answer 4

Idk how lol

Answer 5

\[2\pi \cos \pi x\]
this is what i got

Answer 6

but the answer is
\[2\pi^2 x \cos(\pi x)^2\]

Answer 7

chain rule, is that what you did?

Answer 8

\[\sin^2(\pi x) = \sin(\pi x)^2\]

Answer 9

\(\dfrac{d}{dx} \sin(πx)^2\)

\(= \cos(πx)^2 \dfrac{d ((πx)^2}{dx} \)

\(= \cos(πx)^2 2 (\pi x) \dfrac{d (πx)
}{dx} \)

\(= \cos(πx)^2 2 (\pi x) π\)

\(= 2 \pi^2 x \cos(πx)^2 \)

Chain Rule

Answer 10

hmm.
\[\sin( \pi x)^2 = \sin^2(\pi x) = (\sin(\pi x))^2\]

Answer 11

Whoa, are you supposed to use the power rule for this..??

Answer 12

because how i started it was
\[y'=2\cos(\pi x) \]

Answer 13

then i used the chain rule to multiply it by π

Answer 14

do we agree that

\(y=\sin(πx)^2\)

\(=\sin u \) where \(u = (πx)^2 \)

Answer 15

Yea

Answer 16

so \(\dfrac{dy}{du} = cos u\)

Answer 17

hmm

Answer 18

okay so when you take the derivative of some type of trig function and even if it's raised to a power, you just take the derivative of the trig and don't do anything with the power?

Answer 19

not sure what you mean but....

\(\dfrac{d}{dz} \sin z = \cos z\)

\(\dfrac{d}{dz} \sin^2 z = 2 \sin z \cos z\)

\(\dfrac{d}{dz} \cos^4 z = 4 \cos^3 z \dfrac{d}{dz} (\cos z) = - 4 \cos^3 z \sin z\)

Answer 20

it is \[( \sin(\pi x) )^2\]
apply the power rule alon the chain rule \[2(\sin(\pi x)) \ \cdot d' \sin(\pi x) \cdot d' \pi x\]

Answer 21

\[y= (\sin(\pi x ))^2\]
\[\ y'= 2 ( \sin(\pi x) 0* D~of~ \sin(\pi x) * D~of \pi x\]
\[y' =2 \sin(\pi x) * \cos (\pi x) * \pi \]

Answer 22

ignore 0

Answer 23

lol!! try reading the thread Shy & Sensitive!!

Answer 24

okay so the only thing im confused on is if sin(πx)^2 would be 2cos(πx)(π) or just cos(πx)(π) ??

Answer 25

it really depends upon what you are looking at!

Answer 26

read what ?

Answer 27

For this problem what would it be?

Answer 28

if

\(y = sin (\pi x)^2\)

then we can define \(v = (\pi x)^2\)

so \(y = \sin v\)

and

\(\dfrac{dy}{dv} = \cos v\)

and

\(\dfrac{dy}{dx} = \dfrac{dy}{dv}\bullet \dfrac{dv}{dx}\)

and you can run it from there

Answer 29

so.. cos(πx)^2(π) ?

Answer 30

going tooooooo fast

if \(v = (\pi x)^2\)

then

\(\dfrac{dv}{dx} = 2(\pi x) \pi = 2 \pi^2 x\)

Answer 31

okay i got that part but where does tht come into play in the original equation??

Answer 32

you gotta put it all together

Answer 33

cosv=cos2π^2x ?

Answer 34

waitt.. its
\[cosv=2\pi^2xcos(\pi x) ^2\]

Answer 35

i think

Answer 36

but you're not sure!!

Answer 37

not completely but it sorta makes sense

Answer 38

o.o

Answer 39

\(\color{#0cbb34}{\text{Originally Posted by}}\) @IrishBoy123
it's a real bιτch, for sure
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 40

lol xD yes i agree

Answer 41

I just don't get why you would take the derivative of the inside and multiply it separately if that makes sense

Answer 42

:/

Answer 43

a = b^3

b = c^5

what is \(\dfrac {da}{dc}\) ?!!??

Answer 44

what is that even asking lol derivative of a?

Answer 45

lol!!!

Answer 46

:(((

Answer 47

ok imma forget about this one i'll figure it out tomorrow. I have another one to do lol

Answer 48

@Thatonegirl_
\[y= \sin( \pi x )^2\]
\[y= (\sin( \pi x ))^2\]
power rule + chain rule
\[y=2(\sin( \pi x)) * \cos(\pi x) * \pi\]
not difficult

Answer 49

i don't understand...how can we let v=(pix)^2
it is ( sin(pix)) squared

Answer 50

\[y=x^n\]\[y'=nx^{n-1}\]
\[y=(\sin (\pi x) )^2\]
\[y'=2(\sin(\pi x))^{2-1} * D(\sin(\pi x)) * D(\pi x)\]
D= derivative

Answer 51

\(\color{#0cbb34}{\text{Originally Posted by}}\) @IrishBoy123
lol!! try reading the thread Shy & Sensitive!!
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 52

Originally Posted by @owl
read what ?
End of Quote:
answer my Question.

Answer 53

I'll let you guys know how it goes tomorrow

Answer 54

sure :)