Instantaneous rate of change and deriving a fraction

[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?

Homework Equations

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.

 

[scottdave]

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.

 

[SammyS]

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?

Homework Equations

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.​

.

 

[Mark44]

@Specter, in the future, please post questions about derivatives and integrals in the Calculus & Beyond section. I have moved your thread.

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

Absolutely.

 

[YoungPhysicist]

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

 

[SammyS]

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.

 

[YoungPhysicist]

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” ?

 

[SammyS]

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]

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.

 

[scottdave]

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

 

[Specter]

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?

 

[Specter]

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!

 

[Mark44]

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]

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?

 

[Mark44]

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

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

No, not expand -- just write it as $\frac{-12}{(x + 1)^2}$

 

[SammyS]

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

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.

 

[Mark44]

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]

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.

 

[Mark44]

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.

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

$x+1=2$

$x=1$

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]

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!

 

[Mark44]

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.

 

[Specter]

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.

 

[alan2]

You mean differentiate, not derive.

 

[Specter]

You mean differentiate, not derive.

Whoops, yeah.