# First and second derivatives. Please shed some light

• chubbyorphan
In summary, the conversation is about finding the first and second derivatives of the function g(x) = (2x - 3)/(x + 4). The solution provided involves using the product and quotient rules to arrive at the first and second derivatives. There is some discussion about the accuracy of the solution and the use of a common denominator in the first derivative. The conversation also briefly touches on the concept of taking the derivative of a constant term.
chubbyorphan

## Homework Statement

Hey forum, hope everyone is having a good day!
If someone could check this question out for me that'd be great!

Determine the first and second derivative:

g(x) = (2x - 3)/(x + 4)

## The Attempt at a Solution

g(x) = (2x – 3)(x + 4)^(-1)
g’(x) = (2x – 3)’(x + 4)^(-1) + (2x – 3)[(x + 4)^(-1)]’
g’(x) = (2)(x + 4)^(-1) + (2x – 3)(-1)(x+4)^(-2)(1)
g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

g’’(x) = [2(x + 4)^(-1)]’ – (2x – 3)’(x + 4)^(-2) + (2x – 3)[(x + 4)^(-2)]’
g’’(x) = (-1)2(x + 4)^(-2)(1) – (2)(x + 4)^(-2) + (2x – 3)(-2)(x + 4)^(-3)(1)
g’’(x) = -2(x + 4)^(-2) – 2(x + 4)^(-2) +(–4x + 6)(x + 4)^(-3)

Is this right?
Also should I be dividing my entire final answer by -2

if so it should be:

g''(x) = (x+4)^(-2) + (x + 4)^(-2) + (2x + 3)(x+4)^(-3)

## Homework Equations

also.. when using an online derivative calculator.. it told me that the first derivative for g(x) is g'(x) = 11/(x^2+8x+16) which initially appears much different than my answer g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

However! when I put each first derivative into my graphing calculator the same graph appeared.. implying they are equivalent.. can someone PLEASEEEE explain to me how this is possible or why this makes sense..

***BONUS

for the same question I had two more functions to find the first and second derivative for:

a) y = (3x + 2)^2
y’ = 2(3x + 2)^1(3)
y’ = 6(3x + 2)
y’ = 18x + 12

y’’ = 18

b) f(x) = 5x^2 – 2x
f’(x) = 10x – 2

f’’(x) = 10

I'm pretty sure both a) and b) are right.. so I'll get to my question..
both graphs are a concave up parabola.. neither of which go below 0 on the y-axis

I understand that the 1st derivative, related to the graph of the original function..
where the original function was decreasing.. the 1st derivative graph is below 0 on the y-axis.
Where the original function was increasing.. the 1st derivative graph is above 0 on the y-axis

Now my understanding of the 2nd derivative's graph is somewhat shaky.. but from what I know.. I believe it to be..
where the original function is concave up.. the 2nd derivative graph is above 0 on the y-axis
where the original function is concave down.. the 2nd derivative graph is below 0 on the y-axis

So it would make sense that the 2nd derivative graph for both a) and b) are horizontal lines at y=18 and y=10, respectively. Because both graphs are never concave down.. and a slanted line would eventually go below 0 on the y-axis.

MY QUESTION.. is why y=18 and y=10 is there some connection from the original graph to these numbers I'm not seeing? or is it as simple as 'thats just what the 2nd derivative works out to be'

Thanks so much to anyone who can shed some light on this!

chubbyorphan said:

also.. when using an online derivative calculator.. it told me that the first derivative for g(x) is g'(x) = 11/(x^2+8x+16) which initially appears much different than my answer g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

However! when I put each first derivative into my graphing calculator the same graph appeared.. implying they are equivalent.. can someone PLEASEEEE explain to me how this is possible or why this makes sense..

They use $\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}$

$(x+4)^2=x^2+8x+16$

where does the 11 come from?

$u'v-uv'=?$

What do u and v represent?

Well, in your case, $u=2x-3$ and $v=x+4$.

Look at this formula again $\left(\frac{u}{v}\right)^′=\frac{u^′v−uv^′}{v^2}$ and at $\left(\frac{2x-3}{x+4}\right)'$.

oh wow okay.. I just worked my answer of g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)
through and got the same answer of g'(x) = 11/(x^2+8x+16)

I see what you mean!
so is the second answer (in red) better than my other answer? obviously they are equivalent but you can see I got mine from using the product rule

Ill definitely keep you're rule in mind! AHA the quotient rule I just looked it up! wicked..
but yeah which should I use for my final answer? :S

Also any thoughts on my second derivative answer?

chubbyorphan said:

## Homework Statement

Hey forum, hope everyone is having a good day!
If someone could check this question out for me that'd be great!

Determine the first and second derivative:

g(x) = (2x - 3)/(x + 4)

## The Attempt at a Solution

g(x) = (2x – 3)(x + 4)^(-1)
g’(x) = (2x – 3)’(x + 4)^(-1) + (2x – 3)[(x + 4)^(-1)]’
g’(x) = (2)(x + 4)^(-1) + (2x – 3)(-1)(x+4)^(-2)(1)
g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

g’’(x) = [2(x + 4)^(-1)]’ – (2x – 3)’(x + 4)^(-2) + (2x – 3)[(x + 4)^(-2)]’
g’’(x) = (-1)2(x + 4)^(-2)(1) – (2)(x + 4)^(-2) + (2x – 3)(-2)(x + 4)^(-3)(1)
g’’(x) = -2(x + 4)^(-2) – 2(x + 4)^(-2) +(–4x + 6)(x + 4)^(-3)

Is this right?
Also should I be dividing my entire final answer by -2

if so it should be:

g''(x) = (x+4)^(-2) + (x + 4)^(-2) + (2x + 3)(x+4)^(-3)

## Homework Equations

also.. when using an online derivative calculator.. it told me that the first derivative for g(x) is g'(x) = 11/(x^2+8x+16) which initially appears much different than my answer g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)
...
For your first derivative, use a common denominator.

$\displaystyle g'(x)=2(x + 4)^{-1} – (2x – 3)(x + 4)^{-2}=\frac{2}{x + 4}-\frac{2x – 3}{(x + 4)^{2}}=\dots=\frac{11}{(x + 4)^{2}}$

yes thank you SammyS :) so far I'm at caught up with you to this point
so I'm thinking that g'(x) = 11/(x^2+8x+16

g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

yes?

Also any thoughts on my second derivative answer?g’’(x) = -2(x + 4)^(-2) – 2(x + 4)^(-2) +(–4x + 6)(x + 4)^(-3)that was taken from my
g’(x) =2(x + 4)^(-1) – (2x – 3)(x + 4)^(-2)

I'm thinking it might be way off..
what do you think?

even if that's way wrong..
I was a little confused on how to take the derivative from 2(x+4)^-1
just for my knowledge.. is the derivative of 2(x+4)^-1

-2(x+4)^(-2)

im just confused as to what happens with the 2 outside the bracket.. I'm thinking it stays?

You started with a coefficient of 2, and multiplied it by -1, resulting in -2.

yes, so I did it right then?

Yes.

d/dx( 2(x + 4)-1) = -2(x + 4)-2.

Now that you have that $\displaystyle g'(x)=\frac{11}{(x + 4)^{2}}\,,$ why not find g''(x) using that ?

## 1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It is often thought of as the slope of a curve at a given point.

## 2. What is the first derivative?

The first derivative of a function is the derivative of the function itself. It represents the rate of change of the function at a specific point. In other words, it tells us how fast the function is changing at that point.

## 3. How is the first derivative calculated?

The first derivative is calculated using the limit definition of a derivative. This involves taking the limit of the slope of a secant line as the two points on the line get closer and closer together. Alternatively, it can also be calculated using a shortcut method called differentiation, where we use rules and formulas to find the derivative.

## 4. What is the second derivative?

The second derivative of a function is the derivative of the first derivative. It represents the rate of change of the first derivative at a specific point. In other words, it tells us how fast the slope of the function is changing at that point.

## 5. What is the relationship between the first and second derivatives?

The first derivative and second derivative are related in that the second derivative tells us how the first derivative is changing. If the second derivative is positive, it means the slope of the function is increasing, and if the second derivative is negative, it means the slope of the function is decreasing. Additionally, the second derivative can also tell us about the concavity of the function, or whether it is curved upwards or downwards.

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