Question

y=sin^2(pit) find y' can someone help me please

Answer 1

I was thinking to use the chain rule so i rewrote the problem ==> (sin(pit)^2
than [2sin(pit)(cospit)

Answer 2

yes, and using chain rule only, wouldn't you differentiate pi*x too ?
as pi*x is also a function of x.

Answer 3

f=pi*t, g=sin(f), y=g^2

Answer 4

\((\sin^2 (\pi t))' = 2\sin(\pi t)\cos(\pi t)(\pi t)'\)

Answer 5

how do you differentiate two variables such as (pi*t) ?

Answer 6

pi is not a variable, it's a constant.

Answer 7

\(\pi\approx3.14159\)

Answer 8

pi, a constant, can be thrown out of differentiation

Answer 9

so it would go to 0 and i will only have t left?

Answer 10

\((\pi t)' = \pi (t)' = \pi\)

Answer 11

no, you're differentiating with respect to t, so you'll have pi left

Answer 12

as derivative of t w.r.t t is 1

Answer 13

I dont understand how (pit)'=pi

Answer 14

what is derivative of x,
when differentiated with respect to x ?

Answer 15

@Anna818 how would you differentiate (3x) wrt x?

Answer 16

3

Answer 17

yes, since pi is also a constant like 3,
derivative of pi t will be just pi

Answer 18

ohh lol okay i got it thanks guys!

Answer 19

no problem

Answer 20

From Mathematica 9 Home Edition:
\[\frac{\partial \sin ^2(\pi t)}{\partial t}=2 \pi \sin (\pi t) \cos (\pi t) \]

Answer 21

how would i find the second derivative? i keep getting the wrong answer

Answer 22

First derivative = 2pi sin pit cos pit = pi* sin(2pi t)

now try

Answer 23

2cos(pit)(-sinpit)

Answer 24

the angle is 2pi*t , isn't it ?
and where did the pi(which was outside) go ?

Answer 25

(2cospit)(pi) (- sinpit) (pi)?

Answer 26

\((\pi \sin 2\pi t )' = \pi (\cos(2\pi t)) (2\pi t)' = \pi (\cos(2\pi t)) (2\pi ) =...\)

Answer 27

where did the second pi come from next to sin ?

Answer 28

derivative of 2pi*t

2 pi is constant, so it can be thrown out,

= (2 pi) *derivative of t
= 2pi *1
=2pi

Answer 29

so the beginning was 2in(pit) which can be written as (2pit)*2sin?

Answer 30

what is "in" in 2in(pit)
sin ?

Answer 31

I dont understand why ur differentiating (pi2sinpit)'
1st differentiating went like this (2sinpit)(cospit)(pi)

Answer 32

\[y = \sin^2 (\pi t)\]

Answer 33

(2sinpit)(cospit)(pi) = pi * (sin (2pi t))
because
2 sin A cos A = sin 2A

Answer 34

derive using power rule
\[y' = n u^{n-1}du\]
where u = sin^2 pi t

Answer 35

hence
\[n = 2\]
\[u^{n-1} = \sin \pi t\]
du = derivative of sin (pit) which is
\[du = \pi \cos \pi t\]
Hence (by substituting the values from the variables, we'll get
\[y' = 2 \sin \pi t (\pi \cos \pi t)\]

Answer 36

from identity sin2x = 2sinxcosx
we can also say thay
\[y' = \pi \sin(2 \pi t)\]

Answer 37

hope this helps.

Answer 38

I forgot about that
identity

Answer 39

Thanks!

Answer 40

Thanks everyone!

Answer 41

Yttrium agrees with Mathematica 9 Home Edition.

Answer 42

here the answer
http://www.wolframalpha.com/input/?i=sin%5E2%28pit%29

Answer 43

2pisin(pit)cos(pit)