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How to write the derivative of $V(t) = 2t(1-t)$ as a limit (using first principles).

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1

$begingroup$

I'm trying to find the correct way to write the derivative of $V(t) = 2t(1-t)$ as a limit. i.e.

$f'(x) = limhto 0 frac f(x+h) - f(x) h$

When I put the function into the equation I end up with $frac0h$ but but online calculators are giving me different answers.

Cheers

EDIT: I tried my best to follow the above formula but I don't really understand which values I'm supposed to put in. This is the formula. I came up with

$$
V'(t) = lim_tto 0 frac(2t - 2t + h)^2 - (2t - 2t)^2h = frac0h
$$

calculus limits functions derivatives

share|cite|improve this question

edited 5 mins ago

Asaf Karagila♦

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asked 8 hours ago

C.DeightonC.Deighton

1133 bronze badges

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C.Deighton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You'd better first tell us how you end up with $0/h$.
  $endgroup$
  – trisct
  8 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm trying to find the correct way to write the derivative of $V(t) = 2t(1-t)$ as a limit. i.e.

$f'(x) = limhto 0 frac f(x+h) - f(x) h$

When I put the function into the equation I end up with $frac0h$ but but online calculators are giving me different answers.

Cheers

EDIT: I tried my best to follow the above formula but I don't really understand which values I'm supposed to put in. This is the formula. I came up with

$$
V'(t) = lim_tto 0 frac(2t - 2t + h)^2 - (2t - 2t)^2h = frac0h
$$

calculus limits functions derivatives

share|cite|improve this question

edited 5 mins ago

Asaf Karagila♦

314k3434 gold badges452452 silver badges785785 bronze badges

asked 8 hours ago

C.DeightonC.Deighton

1133 bronze badges

New contributor

C.Deighton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You'd better first tell us how you end up with $0/h$.
  $endgroup$
  – trisct
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to find the correct way to write the derivative of $V(t) = 2t(1-t)$ as a limit. i.e.

$f'(x) = limhto 0 frac f(x+h) - f(x) h$

When I put the function into the equation I end up with $frac0h$ but but online calculators are giving me different answers.

Cheers

EDIT: I tried my best to follow the above formula but I don't really understand which values I'm supposed to put in. This is the formula. I came up with

$$
V'(t) = lim_tto 0 frac(2t - 2t + h)^2 - (2t - 2t)^2h = frac0h
$$

calculus limits functions derivatives

share|cite|improve this question

edited 5 mins ago

Asaf Karagila♦

314k3434 gold badges452452 silver badges785785 bronze badges

asked 8 hours ago

C.DeightonC.Deighton

1133 bronze badges

New contributor

C.Deighton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 5 mins ago

Asaf Karagila♦

314k3434 gold badges452452 silver badges785785 bronze badges

asked 8 hours ago

C.DeightonC.Deighton

1133 bronze badges

New contributor

C.Deighton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 5 mins ago

Asaf Karagila♦

314k3434 gold badges452452 silver badges785785 bronze badges

asked 8 hours ago

C.DeightonC.Deighton

1133 bronze badges

New contributor

C.Deighton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 5 mins ago

Asaf Karagila♦

314k3434 gold badges452452 silver badges785785 bronze badges

edited 5 mins ago

Asaf Karagila♦

314k3434 gold badges452452 silver badges785785 bronze badges

asked 8 hours ago

C.DeightonC.Deighton

1133 bronze badges

New contributor

C.Deighton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

C.DeightonC.Deighton

1133 bronze badges

4 Answers
4

active

oldest

votes

3

$begingroup$

Use $$fracV(t+h)-V(t)h=frac2(t+h)-2(t+h)^2-2t+2t^2h$$
simplify and compute the limit for $h$ tends to zero.
It gives
$$frac2t+2h-2t^2-4th-2h^2-2t+2t^2h$$

share|cite|improve this answer

edited 8 hours ago

Sambo

2,78222 gold badges88 silver badges3232 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

85.8k44 gold badges2929 silver badges6969 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the value of v in the first line?
  $endgroup$
  – C.Deighton
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It denotes your function $v(t)$
  $endgroup$
  – Dr. Sonnhard Graubner
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's being done in the first step to get the second line that's the part I'm stuck on?
  $endgroup$
  – C.Deighton
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @C.Deighton Are you confused about the function notation? We are plugging in the value $t+h$ into your function $V$. Since $V(x) = 2x(1-x)$ for every $x$, we find that $V(t+h) = 2(t+h)(1-(t+h)) = 2(t+h) - 2(t+h)^2$.
  $endgroup$
  – Sambo
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Sambo Yeah that what was exactly what I was confused about, thank you so much.
  $endgroup$
  – C.Deighton
  7 hours ago
  

  
  
  
  

 | 
show 2 more comments

2

$begingroup$

Hint:

It's

$$lim_hto0dfrac2(t+h)(1-(t+h))-2t(1-t)h$$

$$=lim_hto0dfrac2h-4ht-2h^2h$$

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

12.2k11 gold badge99 silver badges2828 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

$$requirecancel]V'(t):=lim_hto0frac2(t+h)(1-(t+h))-2t(1-t)h=$$

$$lim_hto0fraccancel2t-cancel2t^2-2th+2h-2ht-2h^2-cancel2t+cancel2t^2h=lim_hto0frac(2-4t)cancel h-2h^cancel2cancel h=2-4t$$

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

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$endgroup$

add a comment | 

0

$begingroup$

The difference quotient is $dfracf(x+h)-f(x)h=dfrac2(t+h)(1-(t+h))-2t(1-2t)h=dfrac 2h(1-h-2t)h=2(1-h-2t)to2-4t$.

This matches what we get if we just take the derivative: $f'(t)=2(1-t)+2t(-1)=2-4t$.

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

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$endgroup$

add a comment | 

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3

$begingroup$

Use $$fracV(t+h)-V(t)h=frac2(t+h)-2(t+h)^2-2t+2t^2h$$
simplify and compute the limit for $h$ tends to zero.
It gives
$$frac2t+2h-2t^2-4th-2h^2-2t+2t^2h$$

share|cite|improve this answer

edited 8 hours ago

Sambo

2,78222 gold badges88 silver badges3232 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

85.8k44 gold badges2929 silver badges6969 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the value of v in the first line?
  $endgroup$
  – C.Deighton
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It denotes your function $v(t)$
  $endgroup$
  – Dr. Sonnhard Graubner
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's being done in the first step to get the second line that's the part I'm stuck on?
  $endgroup$
  – C.Deighton
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @C.Deighton Are you confused about the function notation? We are plugging in the value $t+h$ into your function $V$. Since $V(x) = 2x(1-x)$ for every $x$, we find that $V(t+h) = 2(t+h)(1-(t+h)) = 2(t+h) - 2(t+h)^2$.
  $endgroup$
  – Sambo
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Sambo Yeah that what was exactly what I was confused about, thank you so much.
  $endgroup$
  – C.Deighton
  7 hours ago
  

  
  
  
  

 | 
show 2 more comments

3

$begingroup$

Use $$fracV(t+h)-V(t)h=frac2(t+h)-2(t+h)^2-2t+2t^2h$$
simplify and compute the limit for $h$ tends to zero.
It gives
$$frac2t+2h-2t^2-4th-2h^2-2t+2t^2h$$

share|cite|improve this answer

edited 8 hours ago

Sambo

2,78222 gold badges88 silver badges3232 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

85.8k44 gold badges2929 silver badges6969 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the value of v in the first line?
  $endgroup$
  – C.Deighton
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It denotes your function $v(t)$
  $endgroup$
  – Dr. Sonnhard Graubner
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What's being done in the first step to get the second line that's the part I'm stuck on?
  $endgroup$
  – C.Deighton
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @C.Deighton Are you confused about the function notation? We are plugging in the value $t+h$ into your function $V$. Since $V(x) = 2x(1-x)$ for every $x$, we find that $V(t+h) = 2(t+h)(1-(t+h)) = 2(t+h) - 2(t+h)^2$.
  $endgroup$
  – Sambo
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Sambo Yeah that what was exactly what I was confused about, thank you so much.
  $endgroup$
  – C.Deighton
  7 hours ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

Use $$fracV(t+h)-V(t)h=frac2(t+h)-2(t+h)^2-2t+2t^2h$$
simplify and compute the limit for $h$ tends to zero.
It gives
$$frac2t+2h-2t^2-4th-2h^2-2t+2t^2h$$

share|cite|improve this answer

edited 8 hours ago

Sambo

2,78222 gold badges88 silver badges3232 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

85.8k44 gold badges2929 silver badges6969 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

Sambo

2,78222 gold badges88 silver badges3232 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

85.8k44 gold badges2929 silver badges6969 bronze badges

edited 8 hours ago

Sambo

2,78222 gold badges88 silver badges3232 bronze badges

edited 8 hours ago

Sambo

2,78222 gold badges88 silver badges3232 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

85.8k44 gold badges2929 silver badges6969 bronze badges

answered 8 hours ago

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

85.8k44 gold badges2929 silver badges6969 bronze badges

2

$begingroup$

Hint:

It's

$$lim_hto0dfrac2(t+h)(1-(t+h))-2t(1-t)h$$

$$=lim_hto0dfrac2h-4ht-2h^2h$$

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

12.2k11 gold badge99 silver badges2828 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Hint:

It's

$$lim_hto0dfrac2(t+h)(1-(t+h))-2t(1-t)h$$

$$=lim_hto0dfrac2h-4ht-2h^2h$$

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

12.2k11 gold badge99 silver badges2828 bronze badges

$endgroup$

add a comment | 

$begingroup$

Hint:

It's

$$lim_hto0dfrac2(t+h)(1-(t+h))-2t(1-t)h$$

$$=lim_hto0dfrac2h-4ht-2h^2h$$

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

12.2k11 gold badge99 silver badges2828 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

J. W. TannerJ. W. Tanner

12.2k11 gold badge99 silver badges2828 bronze badges

answered 8 hours ago

J. W. TannerJ. W. Tanner

12.2k11 gold badge99 silver badges2828 bronze badges

answered 8 hours ago

J. W. TannerJ. W. Tanner

12.2k11 gold badge99 silver badges2828 bronze badges

0

$begingroup$

$$requirecancel]V'(t):=lim_hto0frac2(t+h)(1-(t+h))-2t(1-t)h=$$

$$lim_hto0fraccancel2t-cancel2t^2-2th+2h-2ht-2h^2-cancel2t+cancel2t^2h=lim_hto0frac(2-4t)cancel h-2h^cancel2cancel h=2-4t$$

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

184k1414 gold badges9898 silver badges235235 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

$$requirecancel]V'(t):=lim_hto0frac2(t+h)(1-(t+h))-2t(1-t)h=$$

$$lim_hto0fraccancel2t-cancel2t^2-2th+2h-2ht-2h^2-cancel2t+cancel2t^2h=lim_hto0frac(2-4t)cancel h-2h^cancel2cancel h=2-4t$$

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

184k1414 gold badges9898 silver badges235235 bronze badges

$endgroup$

add a comment | 

$begingroup$

$$requirecancel]V'(t):=lim_hto0frac2(t+h)(1-(t+h))-2t(1-t)h=$$

$$lim_hto0fraccancel2t-cancel2t^2-2th+2h-2ht-2h^2-cancel2t+cancel2t^2h=lim_hto0frac(2-4t)cancel h-2h^cancel2cancel h=2-4t$$

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

184k1414 gold badges9898 silver badges235235 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

184k1414 gold badges9898 silver badges235235 bronze badges

answered 8 hours ago

DonAntonioDonAntonio

184k1414 gold badges9898 silver badges235235 bronze badges

answered 8 hours ago

DonAntonioDonAntonio

184k1414 gold badges9898 silver badges235235 bronze badges

0

$begingroup$

The difference quotient is $dfracf(x+h)-f(x)h=dfrac2(t+h)(1-(t+h))-2t(1-2t)h=dfrac 2h(1-h-2t)h=2(1-h-2t)to2-4t$.

This matches what we get if we just take the derivative: $f'(t)=2(1-t)+2t(-1)=2-4t$.

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

18.2k33 gold badges88 silver badges3131 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

The difference quotient is $dfracf(x+h)-f(x)h=dfrac2(t+h)(1-(t+h))-2t(1-2t)h=dfrac 2h(1-h-2t)h=2(1-h-2t)to2-4t$.

This matches what we get if we just take the derivative: $f'(t)=2(1-t)+2t(-1)=2-4t$.

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

18.2k33 gold badges88 silver badges3131 bronze badges

$endgroup$

add a comment | 

$begingroup$

The difference quotient is $dfracf(x+h)-f(x)h=dfrac2(t+h)(1-(t+h))-2t(1-2t)h=dfrac 2h(1-h-2t)h=2(1-h-2t)to2-4t$.

This matches what we get if we just take the derivative: $f'(t)=2(1-t)+2t(-1)=2-4t$.

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

18.2k33 gold badges88 silver badges3131 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Chris CusterChris Custer

18.2k33 gold badges88 silver badges3131 bronze badges

answered 8 hours ago

Chris CusterChris Custer

18.2k33 gold badges88 silver badges3131 bronze badges

answered 8 hours ago

Chris CusterChris Custer

18.2k33 gold badges88 silver badges3131 bronze badges