# Differentiation Questions

• Feb 13th 2011, 10:54 AM
agent2421
Differentiation Questions

1. f (x) = 7/x^4 - 3/x^3 + 1/x^(1/3)

-28/x^5 + 9/x^4 ... but the third part is killing me... can someoen help me differentiate the lsat part of this question... I know the answer but i need to understand how to get there.

2. f (x) = x^4 + 2x^2 - x + 3 / x

I have no idea what i'm doing wrong in this question... do we treat it using the quotient rule? Can someone show the steps for this as wel.

Thanks :)
• Feb 13th 2011, 11:24 AM
dwsmith
Quote:

Originally Posted by agent2421

1. f (x) = 7/x^4 - 3/x^3 + 1/x^(1/3)

-28/x^5 + 9/x^4 ... but the third part is killing me... can someoen help me differentiate the lsat part of this question... I know the answer but i need to understand how to get there.

2. f (x) = x^4 + 2x^2 - x + 3 / x

I have no idea what i'm doing wrong in this question... do we treat it using the quotient rule? Can someone show the steps for this as wel.

Thanks :)

What would the derivative of $\displaystyle x^{-1/3}=\text{?}$
• Feb 13th 2011, 11:31 AM
agent2421
-1/3 x ^ -4/3

but how do i make it so it becomes 1 / x^4/3

I understand the 4/3 and x goes to the bottom but where does the 1/3 go to?
• Feb 13th 2011, 11:34 AM
dwsmith
Quote:

Originally Posted by agent2421
-1/3 x ^ -4/3

but how do i make it so it becomes 1 / x^4/3

I understand the 4/3 and x goes to the bottom but where does the 1/3 go to?

$\displaystyle \displaystyle\frac{-1}{3}\cdot\frac{1}{x^{4/3}}=\frac{-1\cdot 1}{3\cdot x^{4/3}}=\frac{-1}{3x^{4/3}}$
• Feb 13th 2011, 11:38 AM
agent2421
are you sure about that? The answer in my book says it's just - 1/x^4/3

i guess that's why i was confused otherwise i understood that one. Thanks.

Can you help me with the second one now.
• Feb 13th 2011, 11:41 AM
dwsmith
Quote:

Originally Posted by agent2421
are you sure about that? The answer in my book says it's just - 1/x^4/3

i guess that's why i was confused otherwise i understood that one. Thanks.

Can you help me with the second one now.

The book is wrong. The answer is $\displaystyle \displaystyle -\frac{28}{x^5}+\frac{9}{x^4}-\frac{1}{3x^{4/3}}$

What is so hard about the second one?

What is the derivative of $\displaystyle 3x^{-1}\text{?}$
• Feb 13th 2011, 11:46 AM
agent2421
maybe i stated the question wrong but it 's (x^4+2x^2-x+3) all divided by x. It says solve using the basic rules of differentiation.. can we use hte quotient rule though?
• Feb 13th 2011, 11:48 AM
dwsmith
Quote:

Originally Posted by agent2421
maybe i stated the question wrong but it 's (x^4+2x^2-x+3) all divided by x. It says solve using the basic rules of differentiation.. can we use hte quotient rule though?

You can or you can distribute:

$\displaystyle x^{-1}(x^4+2x^2-x+3)=x^3+2x-1+3x^{-1}$
• Feb 13th 2011, 11:54 AM
agent2421
hm i have no idea what i'm doing wrong... this is what ive done so far by using the quotient rule:

4x^3+4x-1 (1) - 1 (x^4+3x^2-x+3)

which becomes

4x^3 + 4x -1 - x^4 - 2x^2 + x -3

problem is... there's no x^4 in the answer so i know i did something wrong.
• Feb 13th 2011, 11:55 AM
dwsmith
Quote:

Originally Posted by agent2421
hm i have no idea what i'm doing wrong... this is what ive done so far by using the quotient rule:

4x^3+4x-1 (1) - 1 (x^4+3x^2-x+3)

which becomes

4x^3 + 4x -1 - x^4 - 2x^2 + x -3

problem is... there's no x^4 in the answer so i know i did something wrong.

You need to divide by denominator squared.
• Feb 13th 2011, 11:59 AM
agent2421
ok so i get...

-x^4 + 4x^3 - 2x^2 + 5 x - 4 / x^2

and how do i divide it by x^2 again?

does it become:

-x^2 + 4x -2 + (5x/x^2) -4/x^2 ? i'm not sure about the last 2.

thanks for the help, appreciate it.
• Feb 13th 2011, 12:17 PM
dwsmith
Quote:

Originally Posted by agent2421
ok so i get...

-x^4 + 4x^3 - 2x^2 + 5 x - 4 / x^2

and how do i divide it by x^2 again?

does it become:

-x^2 + 4x -2 + (5x/x^2) -4/x^2 ? i'm not sure about the last 2.

thanks for the help, appreciate it.

I don't know why you insisted on the quotient rule since I gave you an easier alternate route but:

$\displaystyle \displaystyle\frac{x(4x^3+4x-1)-(x^4+2x^2-x+3)}{x^2}=\frac{3x^4+2x^2-3}{x^2}=3x^2+2-\frac{3}{x^2}$

Now if you would have taken my advice.

$\displaystyle x^{-1}(x^4+2x^2-x+3)=x^3+2x-1+3x^{-1}$

$\displaystyle \displaystyle 3x^2+2-\frac{3}{x^2}$
• Feb 13th 2011, 12:36 PM
agent2421
Thanks a lot... I just wanted to know how it would work... How do you determine whether to use the quotient rule or the way you did it? If the bottom is only x or x^2 or x^3 will be always do the way you did it ? What if the denominator was 3x^2, would we use the quotient rule then or the way you did it.

Thanks!
• Feb 13th 2011, 12:38 PM
dwsmith
Quote:

Originally Posted by agent2421
Thanks a lot... I just wanted to know how it would work... How do you determine whether to use the quotient rule or the way you did it? If the bottom is only x or x^2 or x^3 will be always do the way you did it ? What if the denominator was 3x^2, would we use the quotient rule then or the way you did it.

Thanks!

You never have to use the quotient rule if you want. Anything in the denominator can be represented as the expression to negative power.