# Math Help - differentiation using quotient rule

1. ## differentiation using quotient rule

Hello, I am stumped on how to differentiate this function:

(1+x) / x^4

My attempt is:
1(x^4) - (1+x)4x^3
= (x^4 - (4x^3+4x^4)) / x^8
= ((-4x^3)-4) / x*

at this point I am not sure if I am headed in the right direction.

Thanks for any help.

2. ## Re: differentiation using quotient rule

Originally Posted by fran1942
Hello, I am stumped on how to differentiate this function:

(1+x) / x^4

My attempt is:
1(x^4) - (1+x)4x^3
= (x^4 - (4x^3+4x^4)) / x^8
= ((-4x^3)-4) / x*

at this point I am not sure if I am headed in the right direction.

Thanks for any help.
$\frac{1+x}{x^4} = \frac{1}{x^4} + \frac{x}{x^4} = x^{-4} + x^{-3}$

now use the power rule

3. ## Re: differentiation using quotient rule

Please make an effort to learn latex - your post is extremely difficult to follow.

We get $f'(x)=\frac{(x^4) - (1+x)4x^3}{x^8}$

$=\frac{(x^4) - 4x^3-4x^4}{x^8}$

$=\frac{-4x^3-3x^4}{x^8}$

Now, as far as I can see, once you divide through a common factor, you're done.

4. ## Re: differentiation using quotient rule

thanks kindly, but I just cant see how I end up with a denominator of x^5 as per the textbook answer.
I thought the common factor and final answer denominator would be x^4 ?

5. ## Re: differentiation using quotient rule

Originally Posted by fran1942
thanks kindly, but I just cant see how I end up with a denominator of x^5 as per the textbook answer.
I thought the common factor and final answer denominator would be x^4 ?
There is a common factor of x^3 in the numerator. When you cancel, you are left with x^5 in the denominator.