Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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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) = ?