How do I use the chain rule to derive certain functions?

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id like some help deriving certain functions using the chain rule
the way our teacher does it is different from what the textbook says
he derives the outermost functions before getting to the innermost functions, this is where
i get confused =(


for example

f(x) = sincos(5x)
i get
f'(x) = (coscos(5x)) (-sin(5x)) (5)
= -5coscos(5x)sin(5x)

im not sure if this is right

another question:

f(x) = sin^4(2x) + cos^4(2x)
 
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here is what i get

f(x) = sin^4(2x) + cos^4(2x)

f'(x) = [ (4sin^3(2x) (cos^4(2x)) (2) ] + [ (4cos^3(2x) (-sin(2x)) (2) ]

= 8sin^3(2x)cos^4(2x) + -8cos^3(2x)sin(2x)


another question - this one i get a different answer than my friend

f(x) = [ ((3x^2 -2)^5) - 1 ] ^5

i get

f(x) = (3x^10 - 2^5 - 1)^5
= 3x^50 - 2^25 - 1^5

f'(x) = 150x^49

here is what I am working on right now - ill post what i get in a few mins

f(x) = |x^1/3|
f(x) = x^2 cos^2(x^2)
f(x) = [ (2x+5)/(7x-9) ]^2/3
 
For the first one in your post you're not applying the chain rule correctly.

d/dx sin(2x)^4 = 4 * sin(2x) ^3 * d/dx (sin2x)

For your second one

(3x^10 - 2^5 - 1)^5 is certainly NOT the same as 3x^50 - 2^25 - 1^5
 
for f(x) = sin^4(2x) + cos^4(2x), i am doing it right I am just deriving it wrong? i should derive each part then add together..if so would the answer be8sin^3(2x)cos(2x) - 8cos^3(2x)sin(2x)

for f(x) = [ ((3x^2 -2)^5) - 1 ] ^5
i get

f'(x) = (150x)(3x^2 - 2)^24another one I am sketched on
f(x) = x^2 cos^2(x^2)

using product rule i get

f'(x) = (2x) (cos^2(2x)) + (x^2) (2cos(x^2)(sin(x^2))(2x)
=2xcos^x(x^2) + 4x^3cos(x^2)sin(x^2)