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**hristo****Member**- Registered: 2005-12-26
- Posts: 8

hi,

how do i solve the following problem:

f (x) = x^5 * 3^x

f' (x) = ?

thanks in advance

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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)

Why did the vector cross the road?

It wanted to be normal.

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**hristo****Member**- Registered: 2005-12-26
- Posts: 8

hmm,

the answer in the book is

f' (x) = x^5 * 3^x *(5/x + ln 3)

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

x^4*3^x(5+x*ln 3)=x^4*3^x*x(5/x+ln3)=x^5*3^x*(5/x+ln3)

IPBLE: Increasing Performance By Lowering Expectations.

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**hristo****Member**- Registered: 2005-12-26
- Posts: 8

omg i can't believe that i didn't think about that. sorry for doubting the answer

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**hristo****Member**- Registered: 2005-12-26
- Posts: 8

and i have another problem which i can't solve:

y = (cos x )^0.7x

thanks

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

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.

*Last edited by mikau (2005-12-26 19:35:14)*

A logarithm is just a misspelled algorithm.

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**Flowers4Carlos****Member**- Registered: 2005-08-25
- Posts: 106

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!

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

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.

A logarithm is just a misspelled algorithm.

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

5 x

f(x)= x 3

4 x 5

¹(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

*Last edited by seerj (2005-12-27 22:36:00)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

hristo, where are you from?

IPBLE: Increasing Performance By Lowering Expectations.

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**seerj****Member**- Registered: 2005-12-21
- Posts: 42

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?

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

Yes. And i'm asking because his name is like someone from East Europe.

IPBLE: Increasing Performance By Lowering Expectations.

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