Understanding the Quotient Rule for Differentiation

[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)

could you please help.

 

[tiny-tim]

Hi morbello!

(try using the X² tag just above the Reply box )

i know 16+x^2 = 2x

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

You mean (16+x²)' = 2x.

And (8(5+x))' = 8(5+x)' = … ?

 

[morbello]

Hi morbello!

(try using the X² tag just above the Reply box )


You mean (16+x²)' = 2x.

And (8(5+x))' = 8(5+x)' = … ?

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

 

[HallsofIvy]

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

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

 

[n!kofeyn]

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

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.

 

[tiny-tim]

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

uh??

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

 

[morbello]

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.

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

 

[n!kofeyn]

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

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.

 

[morbello]

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.

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

 

[tiny-tim]

… on my differentation table it has f(x)=x and f'(x)=1

That's correct!

So (8(x + 5))' = ?​

 

[morbello]

That's correct!

So (8(x + 5))' = ?​

its the same i think ,thank you for helping.

 

[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.

 

[n!kofeyn]

its the same i think ,thank you for helping.

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.

 

[morbello]

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.

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

 

[Cyosis]

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

 

[tiny-tim]

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

morbello, let's take this slowly …

yes, (x)' = 1 … so what's (8x)' ?​

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

 

[morbello]

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

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

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

40+8x=8 could you explain.

 

[Cyosis]

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

 

[morbello]

morbello, let's take this slowly …

yes, (x)' = 1 … so what's (8x)' ?​

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

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

 

[Cyosis]

Why would only the 8x be used?

 

[tiny-tim]

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

Yes, that's right …

(8x)' = (8)' times x plus 8 times (x)'

= 0 times x plus 8 times 1 ​

 

[morbello]

Yes, that's right …

(8x)' = (8)' times x plus 8 times (x)'

= 0 times x plus 8 times 1 ​

ok thank you for your help i have another question but ill leave any questions for today.:)

 

[Mark44]

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

You have written a number of things that aren't true, with the above being one of them. It's not true that 8x = 8.

It is true that d/dx(8x) = 8.

Or to use the notation that several others have been using in this thread, (8x)' = 8.

You have to start getting in the habit of indicating what you are doing, which is differentiating some expression. If you omit the symbols that indicate you are differentiating (but haven't done it yet), people will have a very hard time trying to understand what you're saying.

 

[morbello]

You have written a number of things that aren't true, with the above being one of them. It's not true that 8x = 8.

It is true that d/dx(8x) = 8.

Or to use the notation that several others have been using in this thread, (8x)' = 8.

You have to start getting in the habit of indicating what you are doing, which is differentiating some expression. If you omit the symbols that indicate you are differentiating (but haven't done it yet), people will have a very hard time trying to understand what you're saying.

ok i understand you and will do it in future.