Hello, I'd like to know whether my answer is correct.

My final answer is -26+2h as h approaches 0 or simply -26 because -26+2(0) is -26.

\(2(-5)^{2}-6(-5)+5

=2(25)+30+5

=85\)

So my \(x_{1}\) is -5, while my \(y_{1}\) is 85.

For my \(x_{2}\) and \(y_{2}\):

\(x_{2}\)=5+h and

\(y_{2}\) is:

\(2(-5+h)^{2}-6(-5+h)+5

=2(25-10h+h^{2})+30-6h+5

=50-20h+2h^{2}+30-6h+5

=85-26h+2h^{2}\)

So I have (-5, 85) and (-5+h, 85-26h+2h^2).

\(\frac{(85-26h+2h^{2})-(85)}{(-5+h)-(-5)}

=\frac{-26h+2h^{2}}{h}

=\frac{h(-26+2h)}{h}\)

I cancel the h in the numerator and denominator and have the final result:

\(-26+2h\) as h approaches 0.

So safely, I can say that I have -26 as its derivative at point -5?

Thank you very much!

 

Gosh this really is the hard way to do this.

Not sure why you want the derivative (This isn't homework is it?),

Yes, I make the result -26 at x=-5

 

It looks like you have been asked to apply the definition of the derivative, which is fine, but it would have been nice to indicate so if that is the case.

You want to make your work very easy for anyone (like, say, a grader) to follow what you are doing and why you are doing it.

Following what you are doing here is a bit of a nightmare.

So start of by establishing the specific function form you are using

\(
f(x) = 2x^2-6x+5
\)

Then set the stage for what you are going to do

\(
\frac{df(x)}{dx}=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}
\)

and substitute your function into it

\(
\frac{df(x)}{dx}=\lim_{h \rightarrow 0} \frac{[2(x+h)^2-6(x+h)+5]-[2x^2-6x+5]}{h}
\)

Now expand terms in place

\(
\frac{df(x)}{dx}=\lim_{h \rightarrow 0} \frac{[2(x^2+2xh+h^2)-6(x+h)+5]-[2x^2-6x+5]}{h}
\)

collect terms together according to h

\(
\frac{df(x)}{dx}=\lim_{h \rightarrow 0} \frac{2h^2+(4x-6)h}{h}
\)

and simplify

\(
\frac{df(x)}{dx}=\lim_{h \rightarrow 0} 2h+(4x-6)
\)

Take the limit

\(
\frac{df(x)}{dx}=4x-6
\)

and evaluate it at the specific point of interest

\(
\left\ \frac{df(x)}{dx} \right|_{\small{x=-5}}=4(-5)-6=-26
\)

 

If point (-5, 85) has a derivative -26, so what? I mean what will it help us, in what way? I just studied the book but there is no practical explanation. Is rise over run also applicable to -26?

 

Point's don't have derivatives.

Functions do.

A good way to look at is to think of a graph, perhaps the graph of sales v time.

A graph is a curve drawn on a piece of paper or a screen.

That is a geometrical object. If we want to do calculations with this information the graph is less use than a formula.

Say in my example sales = 2.5 time squared or s = 2.5t^2

A function is a formula describing the graph. After calculating the value of the function (formula) at various values of time we can draw (plot) the graph.

Now suppose the managing directors asks if the sales are rising or falling and how fast.

We can look at the graph and see. That is good for humans, but calculating machines need a formula. Further we can directly read the actual value of sales from the graph but not how fast it is changing.

That is where the derivative comes in.

It is actually short for "the derived function"

That is another formula. A formula that tells us how fast the original function is changing.

Your book probably told you that the speed of change is equal to the slope of a straight line that touches the graph at the time of interest. This line is called the tangent and its slope is given by simple geometry/trigonometry your rise over run.

You can attempt to draw this tangent line by hand and measure the slope at lots of instants in time but,

In my example the speed of change is different at every instant in time so a formula to calculate this is good.

WBahn has shown you how to obtain such a formula for the whole graph in his post.

I also said this is the hard way and since mathematicians are basically lazy they have come up with a set of rules to make things easy. This is not normally introduced until late high school or beginning university, depending where you live.

Does this help?

 

If point (-5, 85) of function f(x) = 2x^2-6x+5 has a derivative of -26. What does it tells us?

Thank you very much!

 

Let's say that your f(x) is the position of an object along a track with x being time in seconds and f(x) being distance in meters. To make it more familiar, let's change the variable x to t. We can call a variable whatever we want, so this changes nothing.

Adding in the proper units, the equation becomes

\(
f(t) = 2 \frac{m}{s^2} t^2 - 6 \frac{m}{s} t + 5m
\)

The first derivative tells us the rate at which the function is changing with respect to variable we are taking the derivative with -- so in this case the rate at which position is changing with respect to time. We call this "velocity". Similarly, the second derivative tells us the rate at which velocity is changing with respect to time. We call this "acceleration".

\(
v(t) = \frac{df(t)}{dt} = 4 \frac{m}{s^2} t - 6 \frac{m}{s}
\)

and

\(
a(t) = \frac{dv(t)}{dt} = \frac{df^2(t)}{dt^2} = 4 \frac{m}{s^2}
\)

So this tells us that at t=-5s that the object was located at

f(t=-5s) = 85m

traveling at a velocity of

v(t=-5s) = -26m/s

with an acceleration of

a(t=-5s) = 4m/s^2

This puts us in a position to ask, and answer, a bunch of different questions. For instance, where will the object be when it comes to rest (albeit momentarily)?

If it is at rest, then the velocity of the object will be zero and hence we are looking for the value of t that results in the first derivative being zero.

\(
v(t) = \frac{df(t)}{dt} = 4 \frac{m}{s^2} t - 6 \frac{m}{s} = 0
t = \frac{6 \frac{m}{s}}{4 \frac{m}{s^2}} = 1.5s
\)

At this time, the position is f(t=1.5s) = 0.5m

This also turns out to be the closest that the object gets to f(t) = 0m (our reference line for measuring distance).

 

You rock! Thanks for explaining, Wbahn and of course studiot.

 

You rock! Thanks for explaining, Wbahn and of course studiot.

My pleasure. It's always a pleasure when someone is as eager to learn as you always are.

 

Is the unit also subtracted for velocity's formula?

I mean \(\frac{m}{s^{2}}-\frac{m}{s}\). If so how come it's just meters. Or we just write m/s because it's the unit for the velocity?

Also how did you come up with an \(2\frac{m}{s^{2}}t^{2}-6\frac{m}{s}t+5m\)??? I have noticed that to each decreasing degree, so do the s and t increases. it just happened that t and s on the last degree is 0 so it's not written. (I'm referring to the last term of the function)

 

You treat units like factors and quantities represented by variables have units as well.

The first term has units of distance/time^2 from the explicit units of the coefficient and time^2 from the variable t^2. When multiplied together, you will be left with distance. Similarly, the second term has explicit units of distance/time and will be multiplied by time, again leaving distance. The final term just has units of distance.

To add terms together, they MUST have the same units.

Notice that there is no ambiguity about whether t is in seconds, or minutes, or hours. You can use anything you want, but you have to then do whatever conversions are needed to get the time units to cancel out.

 

So do you mean the the power of t and s should have the same power always so for it to cancel out, leaving distance??

 

Technically, the power of s is negative and so t and s have powers that are additive inverses. But, with the understanding that we are talking about s being in the denominator, then, yes, in this particular example they have to have the same power.

But notice that in the first derivative the time variable, t, has a power of one less than the power of the explicit time unit, s, in the coefficient's denominator. This is so that one time unit survives in the denominator leaving us with overall units of velocity.

Also notice that we had to choose the units for the original equation but that, after that, the units took care of themselves as we found derivatives or used the equations to answer questions.

 

Wouldn't you mind me showing me how you get the first derivative? I have tried but it seems I'm incorrect.

\(f'(t)=\stackrel{lim}{\rightarrow}0\frac {[2\frac{m}{s^{2}}(t+h)^{2}-6\frac{m}{s}(t+h)+5m]-[2\frac{m}{s}t^{2}-6\frac{m}{s}t+5m]}{h}\)

\(f'(t)=\stackrel{lim}{\rightarrow}0\frac {[2\frac{m}{s^{2}}(t^{2}-2th-h^{2})-6\frac{m}{s}(t+h)+5m]-[2\frac{m}{s}t^{2}-6\frac{m}{s}t+5m]}{h}\)

 

Lightfire, I said this was the hard way.

You are trying to calculate your derivatives from first principles.

Is there any good reason for this?

Even in school we normally just do a few to prove the point then move on to a table of 'standard' derivatives.

To save further difficulty

If f(x) is the standard description of the function and f'(x) the derivative

\(f(x) = a{x^n}\)

\(f'(x) = na{x^{n - 1}}\)

example

\(f(x) = 10{x^5}\)


\(f'(x) = 50{x^4}\)

We prove this general formula once and then use it.

There are tables of such useful derivateives available.

 

Lightfire, I said this was the hard way.

You are trying to calculate your derivatives from first principles.

Is there any good reason for this?

Maybe Lightfire is a masochist.

Do you know what the definition of a sadist is?

A sadist is someone who is nice to a masochist.

I guess that makes WBahn a sadist.

 

There really any isn't good reason. I just bought a book and find a topic with derivative and so do some worksheets. But thanks!!!

 

Maybe Lightfire is a masochist.

Do you know what the definition of a sadist is?

A sadist is someone who is nice to a masochist.

I guess that makes WBahn a sadist.

Ouch!!!!!!!!!!!

 

Maybe Lightfire is a masochist.

Do you know what the definition of a sadist is?

A sadist is someone who is nice to a masochist.

I guess that makes WBahn a sadist.

I always figured my behavior tended to be much more masochist -- at least it sure feels that way!

 

Wouldn't you mind me showing me how you get the first derivative? I have tried but it seems I'm incorrect.

\(f'(t)=\stackrel{lim}{\rightarrow}0\frac {[2\frac{m}{s^{2}}(t+h)^{2}-6\frac{m}{s}(t+h)+5m]-[2\frac{m}{s}t^{2}-6\frac{m}{s}t+5m]}{h}\)

\(f'(t)=\stackrel{lim}{\rightarrow}0\frac {[2\frac{m}{s^{2}}(t^{2}-2th-h^{2})-6\frac{m}{s}(t+h)+5m]-[2\frac{m}{s}t^{2}-6\frac{m}{s}t+5m]}{h}\)

I walked through this in Post #3. Take a look at that and come back with which step starts causing problems (i.e., you can't see how I go to that step from the previous step).

BTW: Thank you for keeping the units properly now that we are talking about a specific function with specific units.