Question

How do I find the derivative of...

Answer 1

\[f(x)=\frac{ 5 }{ (2x)^3 }+2cosx\]

Answer 2

@zepdrix
?

Answer 3

A rule of exponents let's us write it like this:
\(\large\rm f(x)=5(2x)^{-3}+2\cos x\)

Answer 4

Differentiating the first term shouldn't be too bad, power rule into chain rule, ya?

Answer 5

\[-15(2x)^{-4}\]

Answer 6

I am horrible with trigonometric identities though, so I'm not sure what to do with the cos part.

Answer 7

Woops:\[\large\rm \color{royalblue}{\left[5(2x)^{-3}\right]'}=-3\cdot5(2x)^{-4}\cdot\color{royalblue}{(2x)'}\]Forgot your chain rule I think.

Answer 8

\[\large\rm =-3\cdot5(2x)^{-4}\cdot(2)\]

Answer 9

I'm a bit unfamiliar with the rule...
So you would take the 2 from (2x) and multiply it by the rest?

Answer 10

You learn these tricks early on:
Power Rule
Product Rule
Quotient Rule
Chain Rule

Chain Rule is the most difficult of these to master, by far.
You'll need to do a lot lot lot of practice problems to feel comfortable with it.

Answer 11

You're always multiplying by the derivative of the `inner function`.
So yes, for this problem, the inner function is 2x.
We have to multiply by the derivative of 2x on outside like that.

Answer 12

Another example:
\(\large\rm 5(x^2+3)^5\)

I apply the power rule to outermost function which is \(\large\rm 5(\qquad\quad)^5\)
giving me \(\large\rm 5\cdot5(\qquad\quad)^4\)
Chain rule tells me that I have to multiply by the derivative of the inner function.\[\large\rm 5\cdot5(\qquad\quad)^4\cdot\left(\qquad\quad\right)'\]Where ( ) is whatever was inside of the outerfunction that we had.
The prime says to take a derivative.
\[\large\rm 5\cdot5(\qquad\quad)^4\cdot (2x+0)\]So I multiplied by the derivative of x^2+3, the inner function.
So the final result is:\(\large\rm 5\cdot5(x^2+3)^4(2x)\)
Just a silly example :O
I hope the empty brackets didn't make it MORE confusing lol

Answer 13

For the trig stuff.. honestly, just memorize them for now.
Sine and cosine are very cyclical.

When you differentiate sine a number of times, it follows this pattern:
\(\large\rm \sin x\quad\to\quad \cos x\quad\to\quad -\sin x\quad\to\quad-\cos x\quad\to\quad\sin x\)

So if you differentiate sin x, you get cos x.
If you differentiate cos x, you get -sin x.
If you differentiate sin x `four times`, you get right back to sin x! :)

Answer 14

Hey I'm back. Sorry, my internet gave out for a bit. It tends to do that every now and then.

Answer 15

And keep in mind that constant coefficients have effect on the differentiation process.
Just carry the 2 along for the ride.\[\large\rm 2\sin x\quad\to\quad 2\cos x\quad\to\quad -2\sin x\quad\to\quad-2\cos x\quad\to\quad2\sin x\]

Answer 16

have no effect*

Answer 17

I see how in your "Another example" you got the 2x from x^2, but why did the 3 turn into a zero?

Answer 18

Differentiation is all about measuring "change".
Something that is `constant` by definition, does not change.

So when I'm taking a derivative, this "dx" represents an immeasurably small change in x.
So when d/dx looks at the 3, he's asking the question "when x changes a very small amount, how much does 3 change?"

And the answer is 0 amount. It's rate of change is 0.

If you prefer, you can think of it in terms of power rule:\[\large\rm \frac{d}{dx}3\cdot\color{orangered}{1}=\frac{d}{dx}3\cdot\color{orangered}{x^0}=3\cdot0x^{-1}=0\]

Answer 19

Recall that anything to the zero power is just 1. \(\large\rm a^0=1,\qquad \forall a\ne0\).
So I used this clever idea to rewrite 1 as x to the 0.

Answer 20

So any number besides 3 would also turn into zero as well?

Answer 21

Yes.
And later on, they might try to trick you and give you "fancy numbers" as I like to call them.

Answer 22

\[\large\rm \left(e^{4\pi}\right)'=0\]There is no variable!
The whole thing is just a constant!

Answer 23

Ah, okay. That seems simple enough.

Answer 24

Polynomials will follow this progression as you differentiate them:\[\large\rm cubic\quad\to\quad square\quad\to\quad linear \quad\to\quad constant\quad\to\quad 0\]

Answer 25

That's just an illustration of the power rule ^
Obviously higher power follow it as well :)

Answer 26

Bahh I gotta run to the grocery store before it closes :O
You got this figured out? XD

Answer 27

I believe so.
Oh quick question. I just convert cosine to -sine right?

Answer 28

If that will take too much time to explain, then don't worry about it. I don't want to be the cause of you not having food. :)

Answer 29

Good! :D And the 2 comes along for the ride of course.\[\large\rm f(x)=5(2x)^{-3}+2\cos x\]\[\large\rm f'(x)=-3\cdot5(2x)^{-4}(2)-2\sin x\]

Answer 30

Thank you so much!

Answer 31

yay team \c:/

Answer 32

You're better at explaining this than my textbook or teacher.

Answer 33

Ehh it's easy to say that when you're getting 1 on 1 attention XD lol
Gotta give teacher just a little bit of slack haha

Answer 34

Well, I have in the past but eh. I guess I just have a different learning style.
Good night! :)