# Differeation by the quotient.

1. Jul 10, 2009

### morbello

im working on an equation using the quotient rule.

this is the equation.

8(5+x)/16+X^2 and i have to find the f'=

i know 16+x^2 = 2x

im not sure about the 8(5+x) it can not be arctan(x)

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 10, 2009

### tiny-tim

Hi morbello!

(try using the X2 tag just above the Reply box )
You mean (16+x2)' = 2x.

And (8(5+x))' = 8(5+x)' = … ?

3. Jul 10, 2009

### morbello

so it is In(1+x) for 8(5+x)

4. Jul 10, 2009

### HallsofIvy

What is "In"? If you mean ln, No, the derivative of 8(5+x) is certainly NOT ln(x). Why would you think so?

5. Jul 10, 2009

### n!kofeyn

You are wondering about the derivative of 8(5+x)? I'm not for sure why you were thinking about arctan or ln. [8(5+x)]'=[40+8x]'=8. The quotient rule is:
$$\left( \frac{f(x)}{g(x)} \right)' = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$$

In your case f(x)=8(5+x) and g(x)=16+x2.

6. Jul 10, 2009

### tiny-tim

uh??

are you thinking that (1/f(x))' = ∫f(x) ? It isn't!

7. Jul 10, 2009

### morbello

yes i understand im looking for the f'(x) would it be as simple as x=1 and f'(x)=1

8. Jul 10, 2009

### n!kofeyn

I am totally confused by your questions and responses (it looks like all of us are). You are going to have to elaborate, otherwise we are just guessing as to what the problem is.

9. Jul 10, 2009

### morbello

it is wrote down that the f(x)=8(5+x) and i am looking for the f'(x)

im thinking it is , as on my differentation table it has f(x)=x and f'(x)=1

10. Jul 10, 2009

### tiny-tim

That's correct!

So (8(x + 5))' = ?​

11. Jul 10, 2009

### morbello

its the same i think ,thank you for helping.

12. Jul 10, 2009

### Cyosis

Don't say it's the same but give us the expression. If you're thinking that the derivative is the same as the derivative of x you're mistaken.

13. Jul 10, 2009

### n!kofeyn

No it's not. I did it for you in post #5. Distribute the 8 and take the derivative. Do you understand how to take simple derivatives (excluding the product and quotient rules)?

I'm sensing some real misunderstanding on how to take derivatives, so if you can post what you are thinking, then it will allow us to help you out. You need to elaborate in your responses, because I am having a hard time understanding where you are coming from. Do you understand how to take derivatives? Do you understand what the quotient rule means? I posted it in post #5.

Last edited: Jul 10, 2009
14. Jul 10, 2009

### morbello

im not sure what you mean i have jumped a'level maths and am studying this but im working around f(x)= 8(5+x) and f'(x)=1

15. Jul 10, 2009

### Cyosis

No that is wrong. Can you distribute the 8 so that the brackets disappear from your expression?

16. Jul 10, 2009

### tiny-tim

morbello, let's take this slowly …

yes, (x)' = 1 … so what's (8x)' ?​

(if you don't know immediately, then use the product rule )

17. Jul 10, 2009

### morbello

[8(5+x)]'=[40+8x]'=8

you mean this. but im still not sure how you got the =8

40+8x=8 could you explain.

18. Jul 10, 2009

### Cyosis

Do you know what the derivative of a constant is, in this case 40? Try to calculate (40)' and (8x)'.

19. Jul 10, 2009

### morbello

i think i follow you it will only be the 8x that will be used in the rest off the question. which would be 8x=8

20. Jul 10, 2009

### Cyosis

Why would only the 8x be used?