Question

find the derivative of [sin^4(a^2x)]

Answer 1

ahh.. chain rule
Like a function hit-list :D
Your expression, let's express it like
\[\huge \left[ \sin \left( a^{2x} \right) \right]^{4}\]
Key to chain rule is to identify the 'outermost' function. What is it?

Answer 2

by outermost, like in this example, it's the
\[x^{4}\]
So how would you differentiate that?

Answer 3

Yes, that's right. So now, we "killed" ;)
the outermost function.
\[\huge 4\left[ \sin \left( a^{2x} \right) \right]^{3}\]
So now move on to the next level...
Which is
\[\huge \sin (a^{2x})\]
What's the outermost function now?

Answer 4

ok...
Well, it doesn't matter what it was for now, let's work with
\[\huge 4\left[ \sin \left( a^{2}x \right) \right]^{3}\]
Does that look better now? :)

Answer 5

No... for many many reasons... patience :D

Answer 6

We got rid of the outermost function
\[x^{4}\]
And we got, for now:\[4\left[ \sin \left( a^{2}x \right) \right]^{3}\]Leave that to one side, we'll get back to it later.
Now deal with the next 'outermost' function.
Which happens to be
sin x.
How would you differentiate it?

Answer 7

whoa there...
differentiate
sin x
for now...

Answer 8

You only touch the outermost functions, slowly working your way into the innermost.

Answer 9

I have a question... did you forget that a is a constant?
;)

Answer 10

Haha, I was not aware

Answer 11

You did say
d/dx
so it should not affect a (this is not entirely true, but accept it for now)

Answer 12

And lol, be patient
differentiate the
"sin"
first
before that thing inside the sin function :P

Answer 13

And when I say differentiate the "sin" first, I mean leave everything else untouched...

Answer 14

so cos(a^2 x)?

Answer 15

Yes, so you "add that to your hitlist" ;)
lol, I just like to think of things that way...
\[\huge 4\left[ \sin \left( a^{2}x \right) \right]^{3}\left[ \cos \left( a^{2}x \right) \right]\]
But we're not done...
After the sin, there is another function inside
\[a^{2}x\]
how would you differentiate this?

Answer 16

is that just 2a?

Answer 17

2a? again, a^2 is just a constant...

Answer 18

Haha sorry. Well, I still can't grasp this xD

Answer 19

Well, what don't you get?

Answer 20

just the a^2x

Answer 21

I guess Ive already drilled in my head that its supposed to be 2a

Answer 22

if it was d/da, then yes, that is what you should do
but it's d/dx, as you said...

Answer 23

is just 1?

Answer 24

What is just 1?

Answer 25

well in relation to x so a^2 (1)

Answer 26

or just a^2

Answer 27

I'm afraid I don't get you...

Answer 28

Haha its fine, Ive always had issues with chain rule, I dont know why

Answer 29

Better fix that.... chain rule is used extensively :)

Answer 30

I know :(

Answer 31

Well, what would I do for? 4[sin(a^2 (x))]3[cos(a^2(x))]I

Answer 32

Well...
So far, you have
\[\huge 4\left[ \sin \left( a^{2}x \right) \right]^{3}\left[ \cos \left( a^{2}x \right) \right]\]
All that's left is for you to multiply that with the derivative (WITH RESPECT TO X)
of
\[\huge a^{2}x\]
which is just...?

Answer 33

I feel like its such a trick question.. Hmmm..

Answer 34

Imagine angels whispering in your ear in heavenly voices...
"a is a constant, a is a constant, a is a constant...."

Answer 35

Well its not 1, 2a, nor a^2

Answer 36

Why not a^2 ?
If it were
9x,
what would be its derivative?

Answer 37

9?

Answer 38

That's right
Now, what's the derivative of
\[\huge 3^{2}x\]
?

Answer 39

I mean 9

Answer 40

Well, which is it?

Answer 41

.. 6x?

Answer 42

You...
applied...
power rule...
to a constant...
come on
\[\huge 9x = 3^{2}x\]
So why would they have different derivatives...?

Answer 43

So by that statement it just remains a^2 x?

Answer 44

considering 'a' is a constant

Answer 45

Ok, let me just clear something up...
The derivative of
kx
where k is a constant
is
k
Regardless of how weird a constant k is.
It can be as simple as
c
or as complicated as
\[\frac{\ln c + \sin^{2}c}{\cos (c^{2} + 4c + 9)}\]
It doesn't matter, as long as it's a CONSTANT.
In a nutshell:
THE DERIVATIVE OF A CONSTANT TIMES x IS THAT CONSTANT.

Answer 46

What's a constant?
For now, treat as a constant anything that does not involve x, or a function of x.

Answer 47

Its a variable that always retains the same value no matter what, correct?

Answer 48

The mere fact that it's a constant means it's not a variable...

Answer 49

Haha I'm sorry, I understand that, just my words got mixed up :#

Answer 50

Right, so knowing that, is the derivative of
\[a^{2}x\]
clear to you now?

Answer 51

So if its just the constant, a^2 x = a?

Answer 52

If not, I'll accept the fact I'm a complete derp

Answer 53

Ok, I said that the derivative of
kx
is
k.
basically, I just dropped the x.
Please do just that, even in the case of
\[a^{2}x\]

Answer 54

so its just a^2?

Answer 55

YES

Answer 56

derpderpderp

Answer 57

Well, do you understand now?

Answer 58

no, do I put that in the sin or cos part?

Answer 59

now*

Answer 60

no, you've already dealt with that by differentiating the sin into cos earlier, that's why you go further in...

Answer 61

I meant, first you differentiated the
^4 part
And then you differentiated the
sin part
finally, you differentiate the
a^2 x part

Answer 62

And thats just a^2

Answer 63

Yes, and you multiply that to
\[\huge 4\left[ \sin \left( a^{2}x \right) \right]^{3}\left[ \cos \left( a^{2}x \right) \right]\]
And you'll be done.

Answer 64

Thank you very much :)
Haha, and thanks for putting up with my.. ignorance

Answer 65

No problem. :)