Multiple Choice-Derivatives-Please Check?

[Justabeginner]

Homework Statement

Which of the following is not the same as the derivative of y with respect to x, if y= f(x)?

(a) lim as Δx -> 0 of \frac{Δy}{Δx}
(b) lim as h--> 0 of \frac {f(x+h) -f(x)}{h}
(c) lim as x_1 -> x of \frac {f(x_1)-f(x)}{(x_1-x)}
(d) lim as h-> 0 of \frac {f(x) - f(x-h)}{h}
(e) lim as Δx-> 0 of \frac {f(Δx)}{Δx}
(f) lim as h-> 0 of \frac{f(x)- f(x+h)}{-h}

Homework Equations

The Attempt at a Solution

I think C is the correct answer. I immediately ruled out a, b, and e by recognizing them as accurate forms. I thought F was wrong at first, but then I noticed that the original form was changed (negative sign), so that is why the h is negative in the denominator. As for D, I just thought it didn't seem right at all. To be honest, I thought D and C are both incorrect, but the question seems to be asking me for one incorrect answer. Thank you.

 

[1MileCrash]

For C, I suggest you observe what happens if you let x1 = x+h.

For D, I suggest you observe what happens if you let x1=x-h.

I also suggest that you not immediately rule anything out, and try making substitutions and algebraic manipulations. Only one of these is incorrect, and you didn't seem to consider it.

 

[Justabeginner]

Thank you so much for your insight! I wouldn't have thought of it that way. Clearly, D is the right choice then.

 

[Infrared]

No, D is a correct expression for the derivative. Replace h with -h in that expression to see why (the limit is double sided so this is allowed). Take another look at E.

 

[1MileCrash]

Thank you so much for your insight! I wouldn't have thought of it that way. Clearly, D is the right choice then.

No,

For D, let x1=x-h.

 

[Justabeginner]

Okay, I was able to rule out C with your help via algebraic manipulation.
As for A and B, I think that both are standard forms of the derivative, so I will put that aside for now.
For D, I understand now how the -h makes it work.
E: I think I should replace Δx with x + h ?
Then it would make the expression be f(x+h)/(x+h), which I think is wrong.
F: I still think f is correct.

Thank you!

 

[Infrared]

You are right that E is not a correct form for the derivative. Why are you replacing Δx with x+h though?

 

[1MileCrash]

Okay, I was able to rule out C with your help via algebraic manipulation.
As for A and B, I think that both are standard forms of the derivative, so I will put that aside for now.
For D, I understand now how the -h makes it work.
E: I think I should replace Δx with x + h ?
Then it would make the expression be f(x+h)/(x+h), which I think is wrong.
F: I still think f is correct.

Thank you!

For E, instead of a replacement, I suggest you observe the claim it is making.

Basically, I think you'll agree that E hinges on the idea that:

(Delta)f(x) = f((delta)x)

But, remember that delta x is just the difference between two x's. So let's replace delta x with (x1-x2).

Delta f(x) = f(x1-x2)

Or

f(x1)-f(x2) = f(x1-x2)

But, that is merely a property of linear functions, not any function. So E is only true for a very limited class of functions, those with a constant slope.

Sorry for no latex, sent from phone. Hope this helps.

 

[Justabeginner]

Your explanation was so clear. Thank you so much 1MileCrash. I truly appreciate it.

 

[Justabeginner]

You are right that E is not a correct form for the derivative. Why are you replacing Δx with x+h though?

That was a idiotic thing I did, which I only realized after I'd written it. Delta x is the change in x, but I wrote this instead. I do understand the concept of delta x though.

 

[1MileCrash]

You're welcome, good luck. :)

 

[Justabeginner]

Thanks! :)