The derivative of a function such that f(x)=2x^3 would be 6x^2.
What is the derivative of f(x)=1/(2x^3) ?
-1(2x^3)^-2*6x^2
Not sure, but: -3/(2X^4) or (-3/2)(X^-4)
Ah, yes! You’ll see… it’s just the old quotient rule in disguise!
Well, if it’s 1/(2*x^3), then this could be rewritten as 1/2 * x[sup]-3[/sup], whence the derivative is 1/2 * (-3) * x[sup]-4[/sup], or -3/2 * x[sup]-4[/sup], as sleeping wrote.
Incidentally, I find the quotient rule to be a little tedious, and it’s just as easy to use the product rule, which I prefer to do. Here, you don’t even have to do that; it’s just a routine application of the power rule, made a little less obvious because it’s written as a positive exponent, but in the denominator.
The quotient rule is overkill in this case. It’s easier if you think of it as f(x) = 1/2 × x[sup]-3[/sup]. Then it’s pretty easy to see that it’s f’(x) = -3/2 × x[sup]-4[/sup] = -3/(2x[sup]4[/sup]), as sleeping said.
I wish I understood calculus better. Anyone know a good tutorial site? I’ve not found any, despite lengthy searching…all the ones I’ve found seem to be beyond my complete understanding. :mad:
True, the quotient rule is tedious. My friends and I had a cool calculus teacher who encouraged us to use the product rule instead. HOWEVER!
I used to tutor s friend in Calculus, and her prof would not accept this method which she applied to an exam problem! Even though she got the correct answer! What a stupid ass of a math prof!
So, ask your prof (or teacher) if this is ok with him/her before applying as a type of shortcut on an exam! - Jinx
Ah, but of course. Once again we were assuming there was a shortcut rule, in this case for an inverted function… breaking it up and turning X^3 to X^-3 makes it so much easier.
It’s been ten years since I did this. I’m just proud I remembered basic integration and the goddamned Chain Rule, Product Rule, and Quotient Rule. To hell with derivatives, I say.
Thanks!
Honestly, your best bet is going to be to get a calculus book and work through it. Ask questions here; there are many people who will be able to help you.
If that’s the case then buy an inroductory calculus text. A good one will walk you through the subject, simple step by simple step.
A good one is “Calculus” by Varberg and Purcell. You can get a used one pretty cheap from Amazon.
If you in, general, have trouble with math then bypass limit theory and come back to it after you’ve picked up some of the more formulaic stuff.