# How do you differentiate a fraction with x in the numerator and denominator? :/

• Lo.Lee.Ta.
In summary, the fraction confuses the author and causes her to have trouble applying basic algebra. She eventually manages to find the derivative, but is having trouble simplifying it. She gets help from someone who tells her that she needs to learn basic algebra first.
Lo.Lee.Ta.
1. Find derivative:

g(x) = (e^x)/(2 + 3x)2. The fraction confuses me...
I wrote it instead as: (e^x)(3x^-1)(2^-1)

2^(-1) equals .5, so then I thought it was: (e^x)(1.5x^-1)

And then I thought it's time to differentiate...

So I got: (-1.5x^-2)(e^x)

And then I thought, " Since these are multiplied together, do you just differentiate the e^x and it still ends up as (1.5x^-1)(e^x)...?

But none of these ideas are right... =_=

Lo.Lee.Ta. said:
1. Find derivative:

g(x) = (e^x)/(2 + 3x)2. The fraction confuses me...
I wrote it instead as: (e^x)(3x^-1)(2^-1)

2^(-1) equals .5, so then I thought it was: (e^x)(1.5x^-1)

And then I thought it's time to differentiate...

So I got: (1.5x^-1)(e^x)

And then I thought, " Since these are multiplied together, do you just differentiate the e^x and it still ends up as (-1.5x^-2)(e^x)...?

But none of these ideas are right... =_=

You don't have calculus problems above all, you have basic algebra problems. (2+3x)^(-1) means 1/(2+3x). That's not the same as 1/2+1/(3x) or 2^(-1)+(3x)^(-1). Just try using the quotient rule on the given expression.

Unfortunately, basic algebra gives me troubles... =_=

So when I use the product rule, it turns out to be:

(2 + 3x)(e^x) - (e^x)(3x^-1)/ (2 + 3x)^2

And that is:

2e^x + 3xe^x - 3x^-1e^x / 9x^2 + 12x + 4

Can this be simplified any more?

If you tried to move the -3x^-1*e^x to the denominator, would you get:
-3x*e^x or would you get -3x*e^-x?

The reason I ask that is because if it was:

2e^x + 3xe^x / 9x^2 + 12x + 4 - 3xe^x, then wouldn't the 3xe^x in the numerator and denominator cancel, leaving us with a -1?

And then it's: 2e^x -1 / 9x^2 +12x + 4

Basic algebra! HELP!

If you want to use Product Rule, then you can simply try
g(x)=(ex)(2+3x)-1

Since you already have a numerator and denominator arrangement as the function, you can choose to use the Quotient Rule.

Lo.Lee.Ta. said:
Unfortunately, basic algebra gives me troubles... =_=

So when I use the product rule, it turns out to be:

(2 + 3x)(e^x) - (e^x)(3x^-1)/ (2 + 3x)^2

And that is:

2e^x + 3xe^x - 3x^-1e^x / 9x^2 + 12x + 4

Can this be simplified any more?

If you tried to move the -3x^-1*e^x to the denominator, would you get:
-3x*e^x or would you get -3x*e^-x?

You hit a calculus problem before your algebra problems. The derivative of (2+3x) is NOT 3x^(-1). What rule are you trying to apply here?

Lo.Lee.Ta. said:
Unfortunately, basic algebra gives me troubles... =_=

So when I use the product rule, it turns out to be:

(2 + 3x)(e^x) - (e^x)(3x^-1)/ (2 + 3x)^2
No, that's not right, either. You're missing the -1 exponent on the first (2 + 3x) factor.

Lo.Lee.Ta. said:
And that is:

2e^x + 3xe^x - 3x^-1e^x / 9x^2 + 12x + 4
Can this be simplified any more?

If you tried to move the -3x^-1*e^x to the denominator, would you get:
-3x*e^x or would you get -3x*e^-x?

We can show you where you're going wrong, but you're going to need to get yourself up to speed on your own. If you're having problems with basic algebra, and you're trying to do calculus, that's a recipe for disaster.

Lo.Lee.Ta. said:
The reason I ask that is because if it was:

2e^x + 3xe^x / 9x^2 + 12x + 4 - 3xe^x, then wouldn't the 3xe^x in the numerator and denominator cancel, leaving us with a -1?
No. That's not how it works.

Do you think that $$\frac{5 + 1}{2 + 5} = 1/2?$$

The only time you can cancel is when you have the same factors (expressions being multiplied) appearing in the numerator and denominator. You CAN'T cancel if you have the same terms (expressions being added) in numerator and denominator.
Lo.Lee.Ta. said:
And then it's: 2e^x -1 / 9x^2 +12x + 4

Basic algebra! HELP!

Okay. So now I'm just going to try to use the Product Rule.

g(x) = e^x/(2 + 3x)

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

(e^x * -1(2 + 3x)^-2 * 3) + (e^x)(2 + 3x)^-1

e^x * -3(2 + 3x)^-2) + (e^x/2 + 3x)

(-3e^x /(2 + 3x)^2) + (e^x/2 + 3x)

(-3e^x / 9x^2 + 6x + 4) + (e^x)^2/(2 + 3x)^2

(-3e^x/ 9x^2 + 6x + 4) + (e^(2x) / 9x^2 + 6x + 4

= -3e^x + e^(2x) /(9x^2 + 6x +4)

...So this is what I get when I try to follow the product rule...
Is this what the answer is supposed to be?
Am I still doing something wrong? Help.
=_=
Thank you for the help.

For some reason in my earlier posts I said "product rule" when I really meant "quotient rule..."

My method of quotient rule is not right...?

@Dick - Oh, wow. Thanks for pointing out that crucial mistake! :/

When I was trying to do the quotient rule, I said the derivative of of 2 + 3x was 3x^-1!

No, the derivative of that is 3!

So, now it's:

(2 + 3x)(e^x) - (e^x)(3)/(9x^2 + 6x + 4)

2e^x + 3xe^x - 3e^x / (9x^2 + 6x + 4)

= -e^x + 3xe^x / 9x^2 + 6x + 4

...So, that's a different answer from when I did the product rule! Why?!
I am doing something wrong...
Thank you!

OKAY! UGH!

So the computer program I am working on counted the answer: -e^x + 3xe^x/ 9x^2 + 6x + 4 as wrong.

The right answer had to be in the format: -e^x + 3xe^x / (2 +3x)^2
I got the points for the problem, though, so that's good.

BUT I still want to know where I went wrong when I was trying to do the product rule!
Thanks!

Lo.Lee.Ta. said:
OKAY! UGH!

So the computer program I am working on counted the answer: -e^x + 3xe^x/ 9x^2 + 6x + 4 as wrong.

The right answer had to be in the format: -e^x + 3xe^x / (2 +3x)^2
I can't believe that the program thought the above was correct. What you showed here is considered to be
$$-e^x + \frac{3xe^x}{(2 + 3x)^2}$$

The correct answer, in inline text, needs parentheses around the terms in the numerator, like so:
(-e^x + 3xe^x) / (2 +3x)^2
Lo.Lee.Ta. said:
I got the points for the problem, though, so that's good.

BUT I still want to know where I went wrong when I was trying to do the product rule!
Thanks!

Lo.Lee.Ta. said:
@Dick - Oh, wow. Thanks for pointing out that crucial mistake! :/

When I was trying to do the quotient rule, I said the derivative of of 2 + 3x was 3x^-1!

No, the derivative of that is 3!

So, now it's:

(2 + 3x)(e^x) - (e^x)(3)/(9x^2 + 6x + 4)

2e^x + 3xe^x - 3e^x / (9x^2 + 6x + 4)

= -e^x + 3xe^x / 9x^2 + 6x + 4
As already mentioned, you need more parentheses.

This line -- 2e^x + 3xe^x - 3e^x / (9x^2 + 6x + 4) --
should be written as (2e^x + 3xe^x - 3e^x) / (9x^2 + 6x + 4)

This line -- -e^x + 3xe^x / 9x^2 + 6x + 4 --
should be written as (-e^x + 3xe^x) / (9x^2 + 6x + 4)

Also, I don't see any advantage in expanding (3x + 2)2 as you did.

Lo.Lee.Ta. said:
...So, that's a different answer from when I did the product rule! Why?!
I am doing something wrong...
Thank you!

Using the product rule you end up with the sum of two rational expressions. Using the quotient rule, you end up with a single rational expression. In your work with the product rule, get a common denominator and combine the two expressions, and it should simplify to the same as what you got with the quotient rule.

## 1. How do you differentiate a fraction with x in the numerator and denominator?

To differentiate a fraction with x in the numerator and denominator, you will need to use the quotient rule. This rule states that the derivative of a fraction is equal to the denominator multiplied by the derivative of the numerator, minus the numerator multiplied by the derivative of the denominator, all divided by the square of the denominator.

## 2. What is the quotient rule?

The quotient rule is a mathematical formula used to find the derivative of a fraction. It states that the derivative of a fraction is equal to the denominator multiplied by the derivative of the numerator, minus the numerator multiplied by the derivative of the denominator, all divided by the square of the denominator.

## 3. Do I need to simplify the fraction before differentiating?

Yes, it is recommended to simplify the fraction before using the quotient rule to differentiate it. This will make the calculations easier and prevent mistakes.

## 4. Can I use the quotient rule for any fraction?

No, the quotient rule is specifically used for differentiating fractions with variables in both the numerator and denominator. For fractions with only one variable, you can use the power rule.

## 5. Is there an easier way to differentiate fractions?

No, the quotient rule is the most efficient and accurate method for differentiating fractions with variables in both the numerator and denominator. However, with practice, it will become easier and faster to use.

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