derivative for parabola
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I've been revisiting my calculus and have a simple question I can't seem to answer with respect to derivatives of a parabolic function.
Take: $y=x^2$
Derivative $dy = 2x~dx$
However by simply looking at the graph, if you take $x(1,2) = y(1,4)$,
$dfrac{dy}{dx} = dfrac{4-1}{2-1} = 3$ and not $4$ ($2x$ at $x = 2$).
Why is the derivative higher at $4$ versus $dfrac{dy}{dx} = 3$ when looking at the function?
calculus derivatives
edited Nov 16 at 22:23
N. F. Taussig
42.8k93254
asked Nov 16 at 21:59
PEB guy
61
add a comment |
up vote
-2
down vote
favorite
I've been revisiting my calculus and have a simple question I can't seem to answer with respect to derivatives of a parabolic function.
Take: $y=x^2$
Derivative $dy = 2x~dx$
However by simply looking at the graph, if you take $x(1,2) = y(1,4)$,
$dfrac{dy}{dx} = dfrac{4-1}{2-1} = 3$ and not $4$ ($2x$ at $x = 2$).
Why is the derivative higher at $4$ versus $dfrac{dy}{dx} = 3$ when looking at the function?
calculus derivatives
edited Nov 16 at 22:23
N. F. Taussig
42.8k93254
asked Nov 16 at 21:59
PEB guy
61
2
Your question is unclear. You seem to be calculating the slope of the secant line between points $(1, 1)$ and $(2, 4)$. The derivative gives the slope of the tangent line at a point.
– N. F. Taussig
Nov 16 at 22:11
Welcome to MSE, you should use MathJax to write math formulas to make clear and neat.
– hamza boulahia
Nov 16 at 22:23
Welcome to MathSE. This tutorial explains how to use MathJax to typeset mathematics on this site.
– N. F. Taussig
Nov 16 at 22:23
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I've been revisiting my calculus and have a simple question I can't seem to answer with respect to derivatives of a parabolic function.
Take: $y=x^2$
Derivative $dy = 2x~dx$
However by simply looking at the graph, if you take $x(1,2) = y(1,4)$,
$dfrac{dy}{dx} = dfrac{4-1}{2-1} = 3$ and not $4$ ($2x$ at $x = 2$).
Why is the derivative higher at $4$ versus $dfrac{dy}{dx} = 3$ when looking at the function?
calculus derivatives
edited Nov 16 at 22:23
N. F. Taussig
42.8k93254
asked Nov 16 at 21:59
PEB guy
61
I've been revisiting my calculus and have a simple question I can't seem to answer with respect to derivatives of a parabolic function.
Take: $y=x^2$
Derivative $dy = 2x~dx$
However by simply looking at the graph, if you take $x(1,2) = y(1,4)$,
$dfrac{dy}{dx} = dfrac{4-1}{2-1} = 3$ and not $4$ ($2x$ at $x = 2$).
Why is the derivative higher at $4$ versus $dfrac{dy}{dx} = 3$ when looking at the function?
calculus derivatives
calculus derivatives
edited Nov 16 at 22:23
N. F. Taussig
42.8k93254
asked Nov 16 at 21:59
PEB guy
61
edited Nov 16 at 22:23
N. F. Taussig
42.8k93254
asked Nov 16 at 21:59
PEB guy
61
edited Nov 16 at 22:23
N. F. Taussig
42.8k93254
edited Nov 16 at 22:23
N. F. Taussig
42.8k93254
edited Nov 16 at 22:23
N. F. Taussig
42.8k93254
42.8k93254
asked Nov 16 at 21:59
PEB guy
61
asked Nov 16 at 21:59
PEB guy
61
asked Nov 16 at 21:59
PEB guy
61
61
2
Your question is unclear. You seem to be calculating the slope of the secant line between points $(1, 1)$ and $(2, 4)$. The derivative gives the slope of the tangent line at a point.
– N. F. Taussig
Nov 16 at 22:11
Welcome to MSE, you should use MathJax to write math formulas to make clear and neat.
– hamza boulahia
Nov 16 at 22:23
Welcome to MathSE. This tutorial explains how to use MathJax to typeset mathematics on this site.
– N. F. Taussig
Nov 16 at 22:23
add a comment |
2
Your question is unclear. You seem to be calculating the slope of the secant line between points $(1, 1)$ and $(2, 4)$. The derivative gives the slope of the tangent line at a point.
– N. F. Taussig
Nov 16 at 22:11
Welcome to MSE, you should use MathJax to write math formulas to make clear and neat.
– hamza boulahia
Nov 16 at 22:23
Welcome to MathSE. This tutorial explains how to use MathJax to typeset mathematics on this site.
– N. F. Taussig
Nov 16 at 22:23
2
2
Your question is unclear. You seem to be calculating the slope of the secant line between points $(1, 1)$ and $(2, 4)$. The derivative gives the slope of the tangent line at a point.
– N. F. Taussig
Nov 16 at 22:11
Your question is unclear. You seem to be calculating the slope of the secant line between points $(1, 1)$ and $(2, 4)$. The derivative gives the slope of the tangent line at a point.
– N. F. Taussig
Nov 16 at 22:11
Welcome to MSE, you should use MathJax to write math formulas to make clear and neat.
– hamza boulahia
Nov 16 at 22:23
Welcome to MSE, you should use MathJax to write math formulas to make clear and neat.
– hamza boulahia
Nov 16 at 22:23
Welcome to MathSE. This tutorial explains how to use MathJax to typeset mathematics on this site.
– N. F. Taussig
Nov 16 at 22:23
Welcome to MathSE. This tutorial explains how to use MathJax to typeset mathematics on this site.
– N. F. Taussig
Nov 16 at 22:23
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
The derivative is the limit when the distance from your point goes to $0$. So $$frac{dy}{dx}=lim_{hto0}frac{(x+h)^2-x^2}{h}$$
If you choose $h=-1$, you will get a different result than if you have $h=0.1$ or $h=0.01$ or $-0.1$ and $0.01$ on the negative side
answered Nov 16 at 22:32
Andrei
10.1k21025
add a comment |
up vote
0
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3 is the slope of the line connecting $(1,1)$ to $(2,4)$ not the slope of tangent line on parabola in the point $(2,4)$. In fact if we consider the slope of intervening line between $(t,t^2)$ and $(2,4)$ we have$${4-t^2over2}=2+t$$by tending $t$ to 2 we obtain the slope to be 4 as expected.
answered Nov 16 at 22:37
Mostafa Ayaz
12.5k3733
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The derivative is the limit when the distance from your point goes to $0$. So $$frac{dy}{dx}=lim_{hto0}frac{(x+h)^2-x^2}{h}$$
If you choose $h=-1$, you will get a different result than if you have $h=0.1$ or $h=0.01$ or $-0.1$ and $0.01$ on the negative side
answered Nov 16 at 22:32
Andrei
10.1k21025
add a comment |
up vote
1
down vote
The derivative is the limit when the distance from your point goes to $0$. So $$frac{dy}{dx}=lim_{hto0}frac{(x+h)^2-x^2}{h}$$
If you choose $h=-1$, you will get a different result than if you have $h=0.1$ or $h=0.01$ or $-0.1$ and $0.01$ on the negative side
answered Nov 16 at 22:32
Andrei
10.1k21025
add a comment |
up vote
1
down vote
up vote
1
down vote
The derivative is the limit when the distance from your point goes to $0$. So $$frac{dy}{dx}=lim_{hto0}frac{(x+h)^2-x^2}{h}$$
If you choose $h=-1$, you will get a different result than if you have $h=0.1$ or $h=0.01$ or $-0.1$ and $0.01$ on the negative side
answered Nov 16 at 22:32
Andrei
10.1k21025
The derivative is the limit when the distance from your point goes to $0$. So $$frac{dy}{dx}=lim_{hto0}frac{(x+h)^2-x^2}{h}$$
If you choose $h=-1$, you will get a different result than if you have $h=0.1$ or $h=0.01$ or $-0.1$ and $0.01$ on the negative side
answered Nov 16 at 22:32
Andrei
10.1k21025
answered Nov 16 at 22:32
Andrei
10.1k21025
answered Nov 16 at 22:32
Andrei
10.1k21025
answered Nov 16 at 22:32
Andrei
10.1k21025
10.1k21025
add a comment |
add a comment |
up vote
0
down vote
3 is the slope of the line connecting $(1,1)$ to $(2,4)$ not the slope of tangent line on parabola in the point $(2,4)$. In fact if we consider the slope of intervening line between $(t,t^2)$ and $(2,4)$ we have$${4-t^2over2}=2+t$$by tending $t$ to 2 we obtain the slope to be 4 as expected.
answered Nov 16 at 22:37
Mostafa Ayaz
12.5k3733
add a comment |
up vote
0
down vote
3 is the slope of the line connecting $(1,1)$ to $(2,4)$ not the slope of tangent line on parabola in the point $(2,4)$. In fact if we consider the slope of intervening line between $(t,t^2)$ and $(2,4)$ we have$${4-t^2over2}=2+t$$by tending $t$ to 2 we obtain the slope to be 4 as expected.
answered Nov 16 at 22:37
Mostafa Ayaz
12.5k3733
add a comment |
up vote
0
down vote
up vote
0
down vote
3 is the slope of the line connecting $(1,1)$ to $(2,4)$ not the slope of tangent line on parabola in the point $(2,4)$. In fact if we consider the slope of intervening line between $(t,t^2)$ and $(2,4)$ we have$${4-t^2over2}=2+t$$by tending $t$ to 2 we obtain the slope to be 4 as expected.
answered Nov 16 at 22:37
Mostafa Ayaz
12.5k3733
3 is the slope of the line connecting $(1,1)$ to $(2,4)$ not the slope of tangent line on parabola in the point $(2,4)$. In fact if we consider the slope of intervening line between $(t,t^2)$ and $(2,4)$ we have$${4-t^2over2}=2+t$$by tending $t$ to 2 we obtain the slope to be 4 as expected.
answered Nov 16 at 22:37
Mostafa Ayaz
12.5k3733
answered Nov 16 at 22:37
Mostafa Ayaz
12.5k3733
answered Nov 16 at 22:37
Mostafa Ayaz
12.5k3733
answered Nov 16 at 22:37
Mostafa Ayaz
12.5k3733
12.5k3733
add a comment |
add a comment |
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Your question is unclear. You seem to be calculating the slope of the secant line between points $(1, 1)$ and $(2, 4)$. The derivative gives the slope of the tangent line at a point.
– N. F. Taussig
Nov 16 at 22:11
Welcome to MSE, you should use MathJax to write math formulas to make clear and neat.
– hamza boulahia
Nov 16 at 22:23
Welcome to MathSE. This tutorial explains how to use MathJax to typeset mathematics on this site.
– N. F. Taussig
Nov 16 at 22:23