(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. The attempt at a solution

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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# Homework Help: Instantaneous rate of change and deriving a fraction

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