Checking for errors

**the_sensai** · August 19th 2008, 06:09 PM

OKay can i just grap some quick help for a second to just quickly check these answers. Thanks people!.

okay so i got

1) c
2) a
3) a or b differentiation by parts?
4) d
5) b
6) not sure so went for d?
7) b
8) b
9) c

thanks alot for any help people! It really is appreciated.

**Chris L T521** · August 19th 2008, 06:45 PM

Originally Posted by the_sensai

OKay can i just grap some quick help for a second to just quickly check these answers. Thanks people!.

okay so i got

1) c Correct
2) a Incorrect. You need to apply the product rule here.
3) a or b differentiation by parts? Apply the quotient rule. Its A or B. Pick the one that you think is correct.
4) d Incorrect. You need to apply the chain rule. Recall that $\frac{d}{dx}\log_e(u)=\frac{1}{u}\cdot\frac{\,du}{ \,dx}$
5) b Incorrect. Did you apply the first derivative test correctly? Show us you work so we can see where you went wrong.
6) not sure so went for d? Good guess...but what don't you get about the problem? In my opinion, none of them are correct. There's no constant of integration for any of the answers!

7) b Incorrect, but you're really close! What's $\int \cos(x)\,dx$ ??
8) b How did you get this for your answer? Please show us your work so we can see where you went wrong and help you. Did you remember to apply the substitution technique??
9) c Again, please show us your work so we can see where you went wrong and help you. Did you remember to apply the substitution technique??

thanks alot for any help people! It really is appreciated.

I hope this gives you an idea on how you did.

--Chris

**the_sensai** · August 19th 2008, 07:45 PM

okay i messed that up okay!. Gone through it again i will add my workings this time.

So
1) C - correct , simple $7(x)^3$ becomes $21x^2$ then $-4sin(x)$ becomes $-4cos(x) + 5/x$ as loge(x) becomes 5/x.

2) okay so if y= uv then $dy/dx = u dv/dx + v du/dx$

so $u = x^4$ and $v = log_e(x)$

then $du/dx = 4x^3$
and $dv/dx = 1/x$

so it becomes $dy/dx = x^4 . 1/x + log_e(x) . 4x^3$

this then becomes

$4x^3log_e(x) + x^4$

so B?

3) the quotient rule states $dy/dx = (v du/dx - u dv/dx)/v^2$

so then $v = x^2+1 and u = x^2-1$

and $dv/dx = 2x$
$du/dx = 2x$

so then it becomes $(x^2+1.2x - x^2-1.2x)/(x^2+1)^2$

thus $4x/ ( x^2 + 1) ^ 2$

so B?

4) Chain rule!!

my bad so $y = log_e(x^3+3x)$

then $u = (x^3 + 3x)$ $so log_e(u) = 1/u.du/dx$ then

$1/(x^3 + 3x) . 3x^2 + 3$ so becomes

$3(x^2+1)/x^3+3x$

so c?

i'll write the rest whilst i wait for a reply!. Thankyou it has really helped! I'm awful at maths.

**Chris L T521** · August 19th 2008, 07:48 PM

Originally Posted by the_sensai

okay i messed that up okay!. Gone through it again i will add my workings this time.

So
1) C - correct , simple $7(x)^3$ becomes $21x^2$ then $-4sin(x)$ becomes $-4cos(x) + 5/x$ as loge(x) becomes 5/x.

2) okay so if y= uv then $dy/dx = u dv/dx + v du/dx$

so $u = x^4$ and $v = log_e(x)$

then $du/dx = 4x^3$
and $dv/dx = 1/x$

so it becomes $dy/dx = x^4 . 1/x + log_e(x) . 4x^3$

this then becomes

$4x^3log_e(x) + x^4$

so B? Not quite...what's $x^4\cdot\frac{1}{x}$ ??

3) the quotient rule states $dy/dx = (v du/dx - u dv/dx)/v^2$

so then $v = x^2+1 and u = x^2-1$

and $dv/dx = 2x$
$du/dx = 2x$

so then it becomes $(x^2+1.2x - x^2-1.2x)/(x^2+1)^2$

thus $4x/ ( x^2 + 1) ^ 2$

so B? Correct!

4) Chain rule!!

my bad so $y = log_e(x^3+3x)$

then $u = (x^3 + 3x)$ $so log_e(u) = 1/u.du/dx$ then

$1/(x^3 + 3x) . 3x^2 + 3$ so becomes

$3(x^2+1)/x^3+3x$

so c? Correct!

i'll write the rest whilst i wait for a reply!. Thankyou it has really helped! I'm awful at maths.

You got most of them correct so far. Do you understand where you went wrong?

--Chris

**the_sensai** · August 19th 2008, 07:55 PM

yeah thanks, i completely forgot about the rules. I understood when you said. It all makes a bit more sense. Also question 6 do you think i should ask my lecturer about this? Thanks again for your help and taking time to help me!.

**Chris L T521** · August 19th 2008, 07:58 PM

Originally Posted by the_sensai

Also question 6 do you think i should ask my lecturer about this?

Nah. You don't need to. I'm just saying that when you evaluate an indefinite integral, your answer contains a constant of integration, $C$ . None of his answers have the constant of integration. Maybe you could just point that out?

However, one of the choices for that question would be correct...although there's no $C$ .

--Chris

**the_sensai** · August 19th 2008, 07:59 PM

oh and also is $x^4.1/x$ = $x^3$

**Chris L T521** · August 19th 2008, 08:03 PM

Originally Posted by the_sensai

oh and also is $x^4.1/x$ = $x^3$

Yes, so that would make the answer for #2 $4x^3\log_e(x)+x^3$ ...which is choice ___

--Chris

**the_sensai** · August 19th 2008, 08:07 PM

Brilliant!! Thats a massive help. I've never understood the $log_e(x)$ problem. Which you could probably tell!. You've been a massive help. I can't thankyou enough. Cheers!.

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