# Instantaneous rate of change and deriving a fraction

• Specter
In summary: F(x) = (f(x))^{-1} \ \text{and } f(x)=(x+1) \ .##Now use the chain rule to differentiate ##\ F(x) \ .##By the way: The verb form for obtaining a derivative is differentiate, not derive.
Specter

## Homework Statement

Sorry for so many posts lately, hopefully this is allowed.

What tangent points on ##g(x)=\frac {12} {x+1}## has an instantaneous rate of change of -3?

## The Attempt at a Solution

[/B]
I know that once I derive ##g(x)=\frac {12} {x+1}## I can set the derivative to -3 to get my x value then I can substitute the x value into the original equation and find my y. I can't figure out how to derive this.

I know the rules of derivatives but I am very bad at applying them. This is what I have done so far.

##g(x)=\frac {12} {x+1}##

##g'(x)=12 (\frac {1} {x+1})## Removing the constant

## =12((x+1)^{-1})## Applying the power rule

I think I would use the chain rule next because it states that if ##G(x)=(g(x))^{n}## then ##G'(x)=ng'(x)g(x)^{n-1}## which looks like what I have now.

Writing out what is what... ##n = -1##, ##g'(x)=12(\frac {1} {x+1})##, and ##g(x)=\frac {12} {x+1}##

applying it to the chain rule:

##=(-1)(12)(\frac {1} {x+1})(\frac {12} {x+1})^{-2}##

Haven't moved on after this because I'm not confident that this is right.

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You have 1/(x+1) then you have (x+1)^-1 which are equivalent. But then you somehow try to do [1/(x+1)]^-1 which is going to just be x+1. So you need to fix your g(x).
Note you may try the quotient rule as well. If you do everything properly, both approaches will yield the same result.

Specter said:

## Homework Statement

Sorry for so many posts lately, hopefully this is allowed.

What tangent points on ##g(x)=\frac {12} {x+1}## has an instantaneous rate of change of -3?

## The Attempt at a Solution

[/B]
I know that once I derive ##g(x)=\frac {12} {x+1}## I can set the derivative to -3 to get my x value then I can substitute the x value into the original equation and find my y. I can't figure out how to derive this.

I know the rules of derivatives but I am very bad at applying them. This is what I have done so far.

##g(x)=\frac {12} {x+1}##

##g'(x)=12 (\frac {1} {x+1})## Removing the constant

## =12((x+1)^{-1})## Applying the power rule
None of the above three expressions is ##\ g'(x) ##. In fact each is identical to ##\ g(x)##.

What may be useful is to notice that ##\ g(x)\ ## can be written as follows.
##\displaystyle g(x) = 12((x+1)^{-1}) ##​
I think I would use the chain rule next because it states that if ##G(x)=(g(x))^{n}## then ##G'(x)=ng'(x)g(x)^{n-1}## which looks like what I have now.
Your statement of the chain rule is fine, but you will need to state it differently for your problem, because the function ##\ g(x)\ ## is already defined and defined in such a way that it won't fit nicely into that form of the chain rule. Also, the coefficient, 12, makes putting your expression into a nice form for the chain rule a bit complicated.

You could say that ##\displaystyle \ g(x) = 12\cdot F(x) \,, \text{ where } F(x) = (f(x))^{-1} \ \text{and } f(x)=(x+1) \ .##

Now use the chain rule to differentiate ##\ F(x) \ .##
By the way: The verb form for obtaining a derivative is differentiate, not derive.​
.

Specter
SammyS said:
By the way: The verb form for obtaining a derivative is differentiate, not derive.
Absolutely.

Specter and YoungPhysicist
The definition of derivatives is
$$f’(x) = \lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

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Young physicist said:
And by the way, your title “Instantaneous rate of change” does not make sense.A change need time to occur thus cannot happen instantaneously.
No.
The derivative does give the instantaneous rate of change.

SammyS said:
No.
The derivative does give the instantaneous rate of change.
But doesn’t “ instant” means no time? or does it actually mean “almost no time” ?

Young physicist said:
But doesn’t “ instant” means no time? or does it actually mean “almost no time” ?
That's what the limit gives you, the rate of change at an instant. After all, we have such quantities as instantaneous velocity and instantaneous acceleration. Right?

YoungPhysicist
SammyS said:
That's what the limit gives you, the rate of change at an instant. After all, we have such quantities as instantaneous velocity and instantaneous acceleration. Right?
Oops,it seems like my language understanding has a little problem.Previous post edited.

So were you able to solve it? What did you get for the derivative?

SammyS
SammyS said:
None of the above three expressions is ##\ g'(x) ##. In fact each is identical to ##\ g(x)##.

What may be useful is to notice that ##\ g(x)\ ## can be written as follows.
##\displaystyle g(x) = 12((x+1)^{-1}) ##​

Your statement of the chain rule is fine, but you will need to state it differently for your problem, because the function ##\ g(x)\ ## is already defined and defined in such a way that it won't fit nicely into that form of the chain rule. Also, the coefficient, 12, makes putting your expression into a nice form for the chain rule a bit complicated.

You could say that ##\displaystyle \ g(x) = 12\cdot F(x) \,, \text{ where } F(x) = (f(x))^{-1} \ \text{and } f(x)=(x+1) \ .##

Now use the chain rule to differentiate ##\ F(x) \ .##
By the way: The verb form for obtaining a derivative is differentiate, not derive.​
.

It's been a bit since I posted but I haven't had a chance to solve it yet so here's my first attempt. Trying to follow what you said.

##\displaystyle \ g(x) = 12\cdot F(x) \,, \text{ where } F(x) = (f(x))^{-1} \ \text{and } f(x)=(x+1) \ .##

So the chain rule is ## F'(x)=nf'(x)f(x)^{n-1}##

##g'(x)=12((x+1)^{-1})##

##g'(x)=12(-1)(1)(x+1)^{-2}##

Is this correct so far?

scottdave said:
So were you able to solve it? What did you get for the derivative?
Haven't had a chance to until now, hopefully I can figure it out!

Specter said:
It's been a bit since I posted but I haven't had a chance to solve it yet so here's my first attempt. Trying to follow what you said.

##\displaystyle \ g(x) = 12\cdot F(x) \,, \text{ where } F(x) = (f(x))^{-1} \ \text{and } f(x)=(x+1) \ .##

So the chain rule is ## F'(x)=nf'(x)f(x)^{n-1}##

##g'(x)=12((x+1)^{-1})##

##g'(x)=12(-1)(1)(x+1)^{-2}##

Is this correct so far?
Yes, but you definitely should write it in simplest form.

Specter
Mark44 said:
Yes, but you definitely should write it in simplest form.

##g'(x)=-12(x+1)^{-2}##

I'm not sure what to do now, is there more to it? Do I expand?

Specter said:
##g'(x)=-12(x+1)^{-2}##

I'm not sure what to do now, is there more to it? Do I expand?
No, not expand -- just write it as ##\frac{-12}{(x + 1)^2}##

Specter
Specter said:
##g'(x)=-12(x+1)^{-2}##

I'm not sure what to do now, is there more to it? Do I expand?
Now go back to the original problem and continue.

Specter
BTW, for this problem, you could use either the product rule and chain rule (as you did) or the quotient rule. I never recommend the quotient rule when the numerator is just a constant.

Specter and scottdave
SammyS said:
Now go back to the original problem and continue.

I need to set the derivative equal to -3.

##g'(x)=\frac {-12} {(x+1)^{2}} =-3##

##(x+1)^2(\frac {-12} {(x+1)^2}) =-3(x+1)^2##

##-12=-3(x+1)^2##

##\frac {-3(x+1)^2} {-3} = \frac {-12} {-3}##

##\sqrt (x+1)^2 = \sqrt 4##

##x+1=2##

##x=1##

Now I plug the x value into the original equation to find a y value

##g(x)=\frac {12} {x+1}##

##g(1)=\frac {12} {1+1}##

##=\frac {12} {2}##

##=6##

The tangent points (1,6) has an instantaneous rate of change of -3.

Specter said:
I need to set the derivative equal to -3.

##g'(x)=\frac {-12} {(x+1)^{2}} =-3##

##(x+1)^2(\frac {-12} {(x+1)^2}) =-3(x+1)^2##

##-12=-3(x+1)^2##

##\frac {-3(x+1)^2} {-3} = \frac {-12} {-3}##
The step below is only partially correct.
Instead of taking the square root of both sides as you did, solve the quadratic equation ##(x + 1)^2 = 4##. There are two solutions, one of which is x = 1.

Also, in TeX, use braces for the argument of the square root, like this: \sqrt{ (x + 1)^2}. This renders as ##\sqrt{ (x + 1)^2}## so that the vinculum (the bar) extends over everything inside the radical.
Specter said:
##\sqrt (x+1)^2 = \sqrt 4##

##x+1=2##

##x=1##

Now I plug the x value into the original equation to find a y value

##g(x)=\frac {12} {x+1}##

##g(1)=\frac {12} {1+1}##

##=\frac {12} {2}##

##=6##

The tangent points (1,6) has an instantaneous rate of change of -3.

Specter
Mark44 said:
The step below is only partially correct.
Instead of taking the square root of both sides as you did, solve the quadratic equation ##(x + 1)^2 = 4##. There are two solutions, one of which is x = 1.

Also, in TeX, use braces for the argument of the square root, like this: \sqrt{ (x + 1)^2}. This renders as ##\sqrt{ (x + 1)^2}## so that the vinculum (the bar) extends over everything inside the radical.

The two solutions that I got were x=1 and x=-3. How do I know which to use?

I just started using TeX last week so thanks for the tips!

Specter said:
The two solutions that I got were x=1 and x=-3. How do I know which to use?
For this problem, both solutions are applicable. The graph of ##y = \frac{12}{x + 1}## has two points at which the slope of the tangent line is -3.

Last edited:
Mark44 said:
For this problem, both solutions are applicable. The graph of ##y = \frac{12}{x + 1}## has two points at which the slope of the tangent line is -3.

Ah okay. The answer in my review only shows (1,6) so I wasn't sure.

Last edited by a moderator:
You mean differentiate, not derive.

Specter
alan2 said:
You mean differentiate, not derive.
Whoops, yeah.

## What is the definition of instantaneous rate of change?

Instantaneous rate of change is the rate at which a function is changing at a specific point. It is the slope of the tangent line to the curve at that point.

## How is instantaneous rate of change different from average rate of change?

Average rate of change is the average rate at which a function changes over a given interval. Instantaneous rate of change, on the other hand, is the rate at a specific point on the function, which may not be representative of the overall change of the function.

## How do you find the instantaneous rate of change of a function?

The instantaneous rate of change of a function can be found by taking the derivative of the function at a specific point. The derivative is the slope of the tangent line to the curve at that point, which represents the instantaneous rate of change.

## What is the process for deriving a fraction?

To derive a fraction, you need to find the derivative of both the numerator and denominator of the fraction, and then simplify the resulting fraction. This process is also known as the quotient rule.

## Why is understanding instantaneous rate of change and deriving fractions important in science?

Instantaneous rate of change and deriving fractions are important in science because they allow us to analyze and understand how variables are changing in a given system or experiment. This helps us make predictions and draw conclusions about the behavior of the system. It is also essential for understanding more advanced concepts in calculus, which is a fundamental tool in many scientific fields.

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