How do I find the derivative of...

- anonymous

How do I find the derivative of...

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

@zepdrix
?

- zepdrix

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- zepdrix

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

- anonymous

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

- anonymous

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

- zepdrix

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

- zepdrix

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

- anonymous

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

- zepdrix

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.

- zepdrix

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.

- zepdrix

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

- zepdrix

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! :)

- anonymous

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

- zepdrix

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

- zepdrix

have no effect*

- anonymous

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

- zepdrix

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:\[\large\rm \frac{d}{dx}3\cdot\color{orangered}{1}=\frac{d}{dx}3\cdot\color{orangered}{x^0}=3\cdot0x^{-1}=0\]

- zepdrix

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.

- anonymous

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

- zepdrix

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

- zepdrix

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

- anonymous

Ah, okay. That seems simple enough.

- zepdrix

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

- zepdrix

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

- zepdrix

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

- anonymous

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

- anonymous

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. :)

- zepdrix

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

- anonymous

Thank you so much!

- zepdrix

yay team \c:/

- anonymous

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

- zepdrix

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.