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- krassi_holmz
- 2005-12-30 04:19:00
Yes. And i'm asking because his name is like someone from East Europe.
- seerj
- 2005-12-29 21:53:22
krassi_holmz wrote:
hristo, where are you from?
I have a friend called Hristo and he's hungarian. I think that he comes from Hungary. Is it right?
- krassi_holmz
- 2005-12-29 12:40:52
hristo, where are you from?
- seerj
- 2005-12-28 21:34:46
5 x
f(x)= x • 3
4 x 5
ƒ¹(x)= 5x • 3 + x • xln3
And for this:
0.7x
y=(cosx)
I dunno if u have to calculate the derivative, but u can rewrite it in this way
(7/10)x • ln(cosx)
f(x)= e
- mikau
- 2005-12-28 11:54:44
lol. I just saw this yesterday and learned how to do it today. Yeah I suspected you could write it in tersm of e but wasn't sure. Very cool.
- Ricky
- 2005-12-28 01:12:18
d/dx (a^x) = a^x * ln(a)
You can use this, and the chain rule as well. Normally, solving an implicit differentiation (where you don't have y by itself alone on one side of the equation) can lead to some complications later, but in this case, the derivative of ln(y) is 1 / y, so it works out pretty well.
- Flowers4Carlos
- 2005-12-27 18:43:12
it's not that difficult really. just apply log to both sides and differentiate.
y = (cos x )^0.7x
lny = ln(cos x )^0.7x
lny = .7xln(cox)
give it a try! oh and happy holidays everyone!
- mikau
- 2005-12-27 18:30:47
I'm not sure I've done differentiation of the form u^x yet. I think I know how to do it but I'm not sure so I'll let someone who knows tell you.
- hristo
- 2005-12-27 18:11:25
and i have another problem which i can't solve:
y = (cos x )^0.7x
thanks
- hristo
- 2005-12-27 17:54:52
omg i can't believe that i didn't think about that. sorry for doubting the answer
- krassi_holmz
- 2005-12-27 17:31:22
x^4*3^x(5+x*ln 3)=x^4*3^x*x(5/x+ln3)=x^5*3^x*(5/x+ln3)
- hristo
- 2005-12-27 14:30:34
hmm,
the answer in the book is
f' (x) = x^5 * 3^x *(5/x + ln 3)
- mathsyperson
- 2005-12-27 12:24:08
You need the product rule: f'(uv) = f(u)f'(v) + f'(u)f(v)
So, (x^5 * 3^x)' = 3^x(x^5)' + x^5(3^x)'
(x^5)' = 5*x^4, using the basic differentiation of powers rule.
3^x is slightly different, because x is the exponent, so this time the derivative is ln 3* 3^x.
So, your overall derivative is 5 * x^4 * 3^x + x^5 * ln 3 * 3^x
Factorise to make it neater: x^4*3^x(5+x*ln 3)
- hristo
- 2005-12-27 12:18:25
hi,
how do i solve the following problem:
f (x) = x^5 * 3^x
f' (x) = ?
thanks in advance