Basic Derivative Question from Schaum's Outline

[tenni]

Hello,
I don't understand how to work this "solved problem" from chapter 9 of Schaum's Calculus 5th Edition (Ayres, Mendelson). Why are we even doing "y + delta(y)?"
Accepting that it makes sense to do so, I can simplify from 2nd to 3rd line, but nothing else makes any sense. Compare this to problem #1 and 2 from above, which make total sense to me: Why are we doing this differently, and what the heck are we supposed to be doing?

 

[tenni]

I understand it now.

So from comparing to problems 1 and 2 that I included:
change in y = f(x₀ + change in x₀) -f(x₀)

In problem 3, at the start we use:
y + change in y = f(x₀ + change in x₀)

So comparing these two, that means that y =f(x₀)
(Which makes total sense: input x, and it gives you y)

I understand the first two lines of problem 3. Getting from line 3 to 4 makes no sense.

First 2 lines of Problem 3 (I want dy/dx for x³ - x² - 4):
1) y + delta(y) = (x + delta(x))³ - (x + delta(x))² - 4 >>>> because we substitute f(x + delta(x)) for each x in the equation
2) = x³ + 3x²(delta(x)) + 3x(delta(x))² + (delta(x))³ - x² - 2x(delta(x)) - (delta(x))² - 4 >>>> a simple simplification of 1)

Going from 2) to the third line makes no sense.
3) delta(y)) (so somehow we got rid of y on LHS?) = (3x² -2x)(delta(x)) + (3x-1)(delta(x))² + (delta(x))³

(please refer to the screengrab in my initial post if the above is hard to read)

Somehow between 2) and 3), we subtract y from LHS. Meaning, we need to subtract y from RHS. Y = f(x), which is just the original function (right??). So the x³'s cancel out, as do the 4 and x² from RHS of 2). Annd I think typing this out helped me to understand. The terms cancel out accordingly to 3). I didn't understand at first how to combine delta(x)) terms. Then from 3) you just divide each term by delta(x)). It would have been great for him to spell this out, but I understand much better having gone through this. Mods can delete if they like, might be helpful to someone else.

 

[Deleted member 4993]

So from comparing to problems 1 and 2 that I included:
change in y = f(x₀ + change in x₀) -f(x₀)

In problem 3, at the start we use:
y + change in y = f(x₀ + change in x₀)

So comparing these two, that means that y =f(x₀)
(Which makes total sense: input x, and it gives you y)

I understand the first two lines of problem 3. Getting from line 3 to 4 makes no sense.

First 2 lines of Problem 3 (I want dy/dx for x³ - x² - 4):
1) y + delta(y) = (x + delta(x))³ - (x + delta(x))² - 4 >>>> because we substitute f(x + delta(x)) for each x in the equation
2) = x³ + 3x²(delta(x)) + 3x(delta(x))² + (delta(x))³ - x² - 2x(delta(x)) - (delta(x))² - 4 >>>> a simple simplification of 1)

Going from 2) to the third line makes no sense.
3) delta(y)) (so somehow we got rid of y on LHS?) = (3x² -2x)(delta(x)) + (3x-1)(delta(x))² + (delta(x))³

(please refer to the screengrab in my initial post if the above is hard to read)

Somehow between 2) and 3), we subtract y from LHS. Meaning, we need to subtract y from RHS. Y = f(x), which is just the original function (right??). So the x³'s cancel out, as do the 4 and x² from RHS of 2). Annd I think typing this out helped me to understand. The terms cancel out accordingly to 3). I didn't understand at first how to combine delta(x)) terms. Then from 3) you just divide each term by delta(x)). It would have been great for him to spell this out, but I understand much better having gone through this. Mods can delete if they like, might be helpful to someone else.

This is an excellent post - no need to delete it. As you said it surely will help some other student.

 

[tenni]

This is an excellent post - no need to delete it. As you said it surely will help some other student.

Thank you! It is amazing how many times the simple process of trying to articulate my thought process start to finish in a forum post actually makes the answer clear.