Urgent differential help!!

**anthmoo** · October 8th 2007, 01:43 PM

Hey guys

can you help with this please?

9) Differentiate $xe^{\sqrt{x}}$

Please help,

thankyou guys!

**Jhevon** · October 8th 2007, 01:49 PM

Originally Posted by anthmoo

Hey guys

can you help with this please?

9) Differentiate $xe^{\sqrt{x}}$

Please help,

thankyou guys!

we need the product rule here:

recall, by the product rule: $\frac d{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x)$

also recall that $\frac d{dx}e^u = u'e^u$ , where $u$ is a function of $x$

try it and see what you get

**anthmoo** · October 8th 2007, 02:16 PM

I did...but I keep getting 5e^2

I think its differentiating $e^{\sqrt{x}}$ that's the problem. Can you help with that?

**Jhevon** · October 8th 2007, 02:20 PM

Originally Posted by anthmoo

I did...but I keep getting 5e^2

I think its differentiating $e^{\sqrt{x}}$ that's the problem. Can you help with that?

did you see my last post? when finding the derivative of e to some power, you find the derivative of the power and multiply it by e to the old power

example: $\frac d{dx}e^{x^2} = \frac d{dx}x^2 \cdot e^{x^2} = 2xe^{x^2}$

now try $\frac d{dx}e^{\sqrt{x}}$

**anthmoo** · October 8th 2007, 09:53 PM

Thankyou so much!

I can do it now =] all that was needed was that rule you gave for differentiating e to the power of a function. Thank you!

Anthmoo

**Jhevon** · October 8th 2007, 09:53 PM

Originally Posted by anthmoo

Thankyou so much!

I can do it now =] all that was needed was that rule you gave for differentiating e to the power of a function. Thank you!

Anthmoo

so, what's your answer?

**anthmoo** · October 9th 2007, 02:18 AM

I got $2e^2$ ...that's right isn't it?

Thanks =]

**topsquark** · October 9th 2007, 06:34 AM

Originally Posted by anthmoo

I got $2e^2$ ...that's right isn't it?

Thanks =]

No!! The only way you can get a constant after taking a derivative is if the original function is a constant times x. Does $e^{\sqrt{x}}$ look like $cx$ ?

Recall what Jhevon said:

Originally Posted by Jhevon

also recall that $\frac d{dx}e^u = u'e^u$ , where $u$ is a function of $x$

Here we have $u = \sqrt{x}$ . So what is $u^{\prime}$ ?

-Dan

**anthmoo** · October 9th 2007, 08:11 AM

Originally Posted by topsquark

No!! The only way you can get a constant after taking a derivative is if the original function is a constant times x. Does $e^{\sqrt{x}}$ look like $cx$ ?

Recall what Jhevon said:

Here we have $u = \sqrt{x}$ . So what is $u^{\prime}$ ?

-Dan

Wait wait!! I'm sorry, the original question asked "if f'(4)"...if it was f'(4)...would that be right??

Thanks

**topsquark** · October 9th 2007, 08:15 AM

Originally Posted by anthmoo

Wait wait!! I'm sorry, the original question asked "if f'(4)"...if it was f'(4)...would that be right??

Thanks

You tell me.
$u^{\prime} = \frac{1}{2\sqrt{x}}$

-Dan

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