Homework Help
#76
I have a comma problem
Joined: Feb 2009
Location: Fox Lake, IL (from Carrickfergus NI)
Posts: 49,598
#77
BE Forum Addict
Joined: Sep 2010
Location: Maryland (via Belfast, Manchester, Toronto and London)
Posts: 4,802
Re: Homework Help
Here's the mathematical proof:
Differentiation is rate of change, which is basically the slope at a certain point in a graph. A straight line always has the same slope at any point on the line. However, not all graphs are straight lines in which case the slope is dependent on where you are on the graph. The slope at any point would be the slope of a tangent to the point.
Consider 2 points on the non-linear graph y=2x³:
(x, 2x³) and a point nearby (x+h, 2(x+h)³).
The slope of a straight line between the 2 points is the difference in y (delta y) divided by the difference in x (delta x)
delta y = 2(x+h)³ - 2x³ = 2x³ + 6x²h + 6xh² + 2h³ - 2x³ = 6x²h + 6xh² + 2h³
delta x = (x+h) - x = x + h - x = h
dividing delta y (=6x²h + 6xh² + 2h³) by delta x (=h) gives us the slope:
6x² + 6xh + 2h²
Now as h approaches 0 (i.e. the distance between the 2 points gets very small) you get closer to the real tangential slope at the point x.
So h=0, means that our tangential slope at the point x is:
6x² + 6xh + 2h² = 6x² + 6x(0) + 2(0)² = 6x²
Q.E.D.
Differentiation is rate of change, which is basically the slope at a certain point in a graph. A straight line always has the same slope at any point on the line. However, not all graphs are straight lines in which case the slope is dependent on where you are on the graph. The slope at any point would be the slope of a tangent to the point.
Consider 2 points on the non-linear graph y=2x³:
(x, 2x³) and a point nearby (x+h, 2(x+h)³).
The slope of a straight line between the 2 points is the difference in y (delta y) divided by the difference in x (delta x)
delta y = 2(x+h)³ - 2x³ = 2x³ + 6x²h + 6xh² + 2h³ - 2x³ = 6x²h + 6xh² + 2h³
delta x = (x+h) - x = x + h - x = h
dividing delta y (=6x²h + 6xh² + 2h³) by delta x (=h) gives us the slope:
6x² + 6xh + 2h²
Now as h approaches 0 (i.e. the distance between the 2 points gets very small) you get closer to the real tangential slope at the point x.
So h=0, means that our tangential slope at the point x is:
6x² + 6xh + 2h² = 6x² + 6x(0) + 2(0)² = 6x²
Q.E.D.
#79
I have a comma problem
Joined: Feb 2009
Location: Fox Lake, IL (from Carrickfergus NI)
Posts: 49,598
#80
Re: Homework Help
My daughter has been asking for help with her reading comprehension homework. Some of the questions are so vague you can't tell what they are asking for, others seem to ask for information that isn't in the given text. In all cases, she barely has space to write the answers. She is always frustrated by it. It seems to be an exercise in how to deal with complete nonsense - a skill required in most workplaces today.
#82
I have a comma problem
Joined: Feb 2009
Location: Fox Lake, IL (from Carrickfergus NI)
Posts: 49,598
#83
Re: Homework Help
Here's the mathematical proof:
Differentiation is rate of change, which is basically the slope at a certain point in a graph. A straight line always has the same slope at any point on the line. However, not all graphs are straight lines in which case the slope is dependent on where you are on the graph. The slope at any point would be the slope of a tangent to the point.
Consider 2 points on the non-linear graph y=2x³:
Q.E.D.
Differentiation is rate of change, which is basically the slope at a certain point in a graph. A straight line always has the same slope at any point on the line. However, not all graphs are straight lines in which case the slope is dependent on where you are on the graph. The slope at any point would be the slope of a tangent to the point.
Consider 2 points on the non-linear graph y=2x³:
Q.E.D.
WTF, multiply by the index, and drop it one down.
#84
BE Forum Addict
Joined: Sep 2010
Location: Maryland (via Belfast, Manchester, Toronto and London)
Posts: 4,802
Re: Homework Help
Yes, but someone had to prove that rule first! That was done using limits as in the proof I posted. Unfortunately homework help didn't end for me at high school. I'm a physicist, mathematician and computer scientist. So I still get homework help calls from my kids at college.