Question

Derivative of :

Answer 1

...

Answer 2

HEY!! omg i need your help. im going to write it out now

Answer 3

...... are you sure?

Answer 4

\[(1/4 ) \sin^2 2 \theta\]

Answer 5

(1/4) 2(sin(2t)) cos(2t) 2

sin(2t) cos(2t) maybe?

Answer 6

what we have here is pretty much a composite of functions:

C f(g(h(x)))

Answer 7

the constant C pulls out; and we peel off the layers to get to all the parts

Answer 8

C* f'(g(h(x))) * g'(h(x)) * h'(x) * x'

Answer 9

C = 1/4

h' = 2t' = 2

g' = sin(2t)' = cos(2t)

f' = (sin(2t))^2' = 2(sin(2t))

Answer 10

2*2 = 4; /4 = 1; and we are left with sin(2t)cos(2t)

Answer 11

im honestly having issues with these types of problems, i feel bad , im learning this on my own, and im struggling. Idk how these work, and im constantly asking on here :/

Answer 12

i learned these on me own too

Answer 13

think of the chain rule like this if it helps:
\[\frac{df}{dx}=\frac{df}{dg}*\frac{dg}{dh}*\frac{dh}{dx}\]

Answer 14

in order to derive for df/dx; you have to get thru all the derivatives that lead to it

Answer 15

wait what happened in the end?

Answer 16

ohh ok

Answer 17

what happened at what end?

Answer 18

i cant really say that i can keep up with this conversation to well :)

Answer 19

i was lost with the 2*2 /4 =1 that step just came at me and i didnt know what happaned. im still lost to be honest i feel like theres something i dont know. like my next question im looking at in fear

Answer 20

post!

Answer 21

no being scared. just remember the chain rule.

Answer 22

\[y= \sqrt{x} + 1/4 \sin(2x)^2 \]

Answer 23

its also finding the derivative. these type of problems are driving me insane, i feel like im missing some peice of information that is making this seem harder then it is