anonymous
  • anonymous
How do I find the derivative of...
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
\[f(x)=\frac{ 5 }{ (2x)^3 }+2cosx\]
anonymous
  • 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
  • zepdrix
Differentiating the first term shouldn't be too bad, power rule into chain rule, ya?
anonymous
  • anonymous
\[-15(2x)^{-4}\]
anonymous
  • anonymous
I am horrible with trigonometric identities though, so I'm not sure what to do with the cos part.
zepdrix
  • 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
  • zepdrix
\[\large\rm =-3\cdot5(2x)^{-4}\cdot(2)\]
anonymous
  • anonymous
I'm a bit unfamiliar with the rule... So you would take the 2 from (2x) and multiply it by the rest?
zepdrix
  • 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
  • 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
  • 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
  • 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
  • anonymous
Hey I'm back. Sorry, my internet gave out for a bit. It tends to do that every now and then.
zepdrix
  • 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
  • zepdrix
have no effect*
anonymous
  • 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
  • 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
  • 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
  • anonymous
So any number besides 3 would also turn into zero as well?
zepdrix
  • zepdrix
Yes. And later on, they might try to trick you and give you "fancy numbers" as I like to call them.
zepdrix
  • zepdrix
\[\large\rm \left(e^{4\pi}\right)'=0\]There is no variable! The whole thing is just a constant!
anonymous
  • anonymous
Ah, okay. That seems simple enough.
zepdrix
  • 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
  • zepdrix
That's just an illustration of the power rule ^ Obviously higher power follow it as well :)
zepdrix
  • zepdrix
Bahh I gotta run to the grocery store before it closes :O You got this figured out? XD
anonymous
  • anonymous
I believe so. Oh quick question. I just convert cosine to -sine right?
anonymous
  • 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
  • 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
  • anonymous
Thank you so much!
zepdrix
  • zepdrix
yay team \c:/
anonymous
  • anonymous
You're better at explaining this than my textbook or teacher.
zepdrix
  • 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
  • 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.