Differentiation

**xxJaRxx** · December 16th 2009, 11:19 AM

Hi i need to differentiate the following function:

f(x) = e^x . (x+2) . (x-1)

I know the answer is

f'(x) = e^x (x^2 + 3x -1)

But have no idea how to get to this! Any help please?

**xxJaRxx** · December 16th 2009, 11:30 AM

Plus then i need to make f'(x) = 0 and find that x = -3.30 and x = 0.30 :\

**adkinsjr** · December 16th 2009, 11:44 AM

I think the function looks like this:

$f(x)=e^{x(x+2)(x-1)}$

If I'm correct, then the derivative is $f'(x)=e^u\frac{du}{dx}$ where $u=x(x+2)(x-1)$ . This is the chain rule.

It's difficult to read what you typed. You should try to use the latex.

**xxJaRxx** · December 16th 2009, 11:49 AM

No sorry the question is

f(x) = (e^x) . (x+2) . (x-1)

and answer is

f'(x) = (e^x) (x^2 + 3x -1)

**pickslides** · December 16th 2009, 12:04 PM

If the equation is $f(x) = e^x (x+2) (x-1)$

Then $f'(x) =\left[e^x \right]' (x+2) (x-1) +e^x \left[(x+2) (x-1)\right]'$

**xxJaRxx** · December 16th 2009, 01:15 PM

Thanks for everyones help ive got it sorted now!

Need help with this now!

Whats the derivative of:

Sin ( (x^2) + x )

**pickslides** · December 16th 2009, 01:20 PM

Lets say

$y = \sin( x^2 + x )$ and $u = x^2 + x\implies y=\sin(u)$

Now $\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

**xxJaRxx** · December 16th 2009, 01:25 PM

so the answer is..

(2x + 1) . cos( (x^2) + x )

Is that right?

**xxJaRxx** · December 16th 2009, 02:10 PM

Okay so i swear the answer to this question is wrong!

Bit of integration instead of differentiation!

$\int3xsin(7-x) dx$

and the answer is

$3xcos(7-x) + 3sin(7-x) + c$

Whereas I would have said it was

$3xsin(7-x) - 3cos(7-x) + c$

Any help?

**pickslides** · December 16th 2009, 03:15 PM

Originally Posted by xxJaRxx

so the answer is..

(2x + 1) . cos( (x^2) + x )

Is that right?

Good work!

**pickslides** · December 16th 2009, 03:20 PM

Originally Posted by xxJaRxx

Okay so i swear the answer to this question is wrong!

Bit of integration instead of differentiation!

$\int3xsin(7-x) dx$

and the answer is

$3xcos(7-x) + 3sin(7-x) + c$

Whereas I would have said it was

$3xsin(7-x) - 3cos(7-x) + c$

Any help?

Read this Integration by parts - Wikipedia, the free encyclopedia

For $\int uv' = uv - \int vu'$

In your case make $u=3x$ and $v' = \sin(7-x)$

now find $u'$ and $v$

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