# Where am i going wrong on taking the derivative?

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• August 5th 2009, 09:37 AM
Darkhrse99
Where am i going wrong on taking the derivative?
The problem is $\frac{t^8+9t+3}{t^2}$

I used the qoutent rule and got $\frac{t^2(8t^7+9)-t^8+9t+3(2t)}{(t^2)^2}$

I ended with $\frac{3t(2t^8+9t+2)}{t^4}$

This work isn't looking like the answer at all. Where am I going wrong?

Thanks
• August 5th 2009, 09:41 AM
AlephZero
Quote:

Originally Posted by Darkhrse99
The problem is $\frac{t^8+9t=3}{t^2}$

I used the qoutent rule and got $\frac{t^2(8t^7+9)-t^8+9t+3(2t)}{(t^2)^2}$

I ended with $\frac{3t(2t^8+9t+2)}{t^4}$

This work isn't looking like the answer at all. Where am I going wrong?

Thanks

The = in the initial problem is supposed to be a +, I think? Anyway, when you did the Quotient Rule, you forgot to multiply the terms in the numerator (2nd part)....
• August 5th 2009, 09:58 AM
Darkhrse99
The middle work I left out is $(8t^9+9t^2)-2T^9+18t^2+6t$

doing subtraction gives you $\frac{6t^9+27t^2+6t}{(t^2)^2}$

I did the multiplication in the numerator.
• August 5th 2009, 10:12 AM
AlephZero
Quote:

Originally Posted by Darkhrse99
The middle work I left out is $(8t^9+9t^2)-2T^9+18t^2+6t$

doing subtraction gives you $\frac{6t^9+27t^2+6t}{(t^2)^2}$

I did the multiplication in the numerator.

I don't follow that... but what I'm saying is, the Quotient Rule gives

$\frac{t^2(8t^7+9)-(t^8+9t+3)(2t)}{t^4},$

which is not what you had.
• August 5th 2009, 10:16 AM
Jameson
Quote:

Originally Posted by Darkhrse99
$\frac{t^8+9t+3}{t^2}$

I would break it up into three terms, which seems a lot easier to me.

$\frac{t^8}{t^2}+\frac{9t}{t^2}+\frac{3}{t^2}$
• August 5th 2009, 10:43 AM
Darkhrse99
Quote:

Originally Posted by Jameson
I would break it up into three terms, which seems a lot easier to me.

$\frac{t^8}{t^2}+\frac{9t}{t^2}+\frac{3}{t^2}$

Jameson,

Using the qoutent rule, I came up with $\frac{-6t^9+3t^2+6t}{(t^2)^2}$

How can I simplify this more?
• August 5th 2009, 10:50 AM
Darkhrse99
Quote:

Originally Posted by AlephZero
I don't follow that... but what I'm saying is, the Quotient Rule gives

$\frac{t^2(8t^7+9)-(t^8+9t+3)(2t)}{t^4},$

which is not what you had.

I had posted this problem the exact way you did in my first post.
• August 5th 2009, 10:54 AM
Darkhrse99
Quote:

Originally Posted by Darkhrse99
I had posted this problem the exact way you did in my first post.

When you multiply this out I get $\frac{(8t^9+9t^2)-(2t^9+18t^2+6t)}{t^4}$

Is this correct?
• August 5th 2009, 11:09 AM
AlephZero
Quote:

Originally Posted by Darkhrse99
The problem is $\frac{t^8+9t+3}{t^2}$

I used the qoutent rule and got $\frac{t^2(8t^7+9)-t^8+9t+3(2t)}{(t^2)^2}$

I ended with $\frac{3t(2t^8+9t+2)}{t^4}$

This work isn't looking like the answer at all. Where am I going wrong?

Thanks

Quote:

Originally Posted by AlephZero
I don't follow that... but what I'm saying is, the Quotient Rule gives

$\frac{t^2(8t^7+9)-(t^8+9t+3)(2t)}{t^4},$

which is not what you had.

Quote:

Originally Posted by Darkhrse99
I had posted this problem the exact way you did in my first post.

As you can see, you absolutely did not have it posted that way, and that is the beginning of where your calculations went wrong.
• August 5th 2009, 11:12 AM
Darkhrse99
Quote:

Originally Posted by AlephZero
As you can see, you absolutely did not have it posted that way, and that is the beginning of where your calculations went wrong.

You right, I left the bracket's out by mistake and didn't sqaure the $t^2$

Is my math correct though?
• August 5th 2009, 11:20 AM
AlephZero
Quote:

Originally Posted by Darkhrse99
Jameson,

Using the qoutent rule, I came up with $\frac{-6t^9+3t^2+6t}{(t^2)^2}$

How can I simplify this more?

Quote:

Originally Posted by Darkhrse99
You right, I left the bracket's out by mistake and didn't sqaure the $t^2$

Is my math correct though?

If your answer is the one above, then I'd have to say no. If you've revised your answer, then you need to provide it, as I haven't seen you give another.
• August 5th 2009, 11:26 AM
Darkhrse99
Is the answer $6t^5+\frac{9}{t^2}+\frac{6}{t^3}$?
• August 5th 2009, 11:38 AM
skeeter
Quote:

Originally Posted by Darkhrse99
The problem is $\frac{t^8+9t+3}{t^2}$

work smart, not hard.

$\frac{t^8+9t+3}{t^2} = t^6 + 9t^{-1} + 3t^{-2}$

$\frac{d}{dt}\left[t^6 + 9t^{-1} + 3t^{-2}\right] = 6t^5 - 9t^{-2} - 6t^{-3} = 6t^5 - \frac{9}{t^2} - \frac{6}{t^3}$
• August 5th 2009, 11:40 AM
AlephZero
Quote:

Originally Posted by Darkhrse99
Is the answer $6t^5+\frac{9}{t^2}+\frac{6}{t^3}$?

No. You have asked where your calculations went wrong, and I have shown you over and again where it is. As I said, you have not done the multiplication in the numerator correctly, because you've neglected to distribute the -1. So, once again, the numerator is

$t^2(8t^7+9)-(t^8+9t+3)(2t)$

$=8t^9+9t^2-2t^9-18t^2-6t$

$=6t^9-9t^2-6t$.
• August 5th 2009, 11:43 AM
Darkhrse99
Quote:

Originally Posted by skeeter
work smart, not hard.

$\frac{t^8+9t+3}{t^2} = t^6 + 9t^{-1} + 3t^{-2}$

$\frac{d}{dt}\left[t^6 + 9t^{-1} + 3t^{-2}\right] = 6t^5 - 9t^{-2} - 6t^{-3} = 6t^5 - \frac{9}{t^2} - \frac{6}{t^3}$

I believe you ment $6t^5-\frac{9}{t^2}-\frac{6}{t^3}$?
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