Mixing
Differentiation from first principles
$begingroup$
How would I prove that
$$frac d{dt}(1+t-2t^2) = 1-4t$$
Using differentiation from first principles.
Then I tried to use the equation:
$$frac{f(t+h)-f(t)}h$$
Is this correct and what do I do after this.
ordinary-differential-equations proof-explanation
edited Jan 15 at 13:19
Shubham Johri
5,177717
asked Jan 15 at 13:08
user8469209user8469209
122
$endgroup$
add a comment |
$begingroup$
How would I prove that
$$frac d{dt}(1+t-2t^2) = 1-4t$$
Using differentiation from first principles.
Then I tried to use the equation:
$$frac{f(t+h)-f(t)}h$$
Is this correct and what do I do after this.
ordinary-differential-equations proof-explanation
edited Jan 15 at 13:19
Shubham Johri
5,177717
asked Jan 15 at 13:08
user8469209user8469209
122
$endgroup$
add a comment |
$begingroup$
How would I prove that
$$frac d{dt}(1+t-2t^2) = 1-4t$$
Using differentiation from first principles.
Then I tried to use the equation:
$$frac{f(t+h)-f(t)}h$$
Is this correct and what do I do after this.
ordinary-differential-equations proof-explanation
edited Jan 15 at 13:19
Shubham Johri
5,177717
asked Jan 15 at 13:08
user8469209user8469209
122
$endgroup$
How would I prove that
$$frac d{dt}(1+t-2t^2) = 1-4t$$
Using differentiation from first principles.
Then I tried to use the equation:
$$frac{f(t+h)-f(t)}h$$
Is this correct and what do I do after this.
ordinary-differential-equations proof-explanation
ordinary-differential-equations proof-explanation
edited Jan 15 at 13:19
Shubham Johri
5,177717
asked Jan 15 at 13:08
user8469209user8469209
122
edited Jan 15 at 13:19
Shubham Johri
5,177717
asked Jan 15 at 13:08
user8469209user8469209
122
edited Jan 15 at 13:19
Shubham Johri
5,177717
edited Jan 15 at 13:19
Shubham Johri
5,177717
edited Jan 15 at 13:19
Shubham Johri
5,177717
5,177717
asked Jan 15 at 13:08
user8469209user8469209
122
asked Jan 15 at 13:08
user8469209user8469209
122
asked Jan 15 at 13:08
user8469209user8469209
122
122
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, that’s the First Principle, or the definition of a derivative:
$$f’(t) = lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
You have $f(t) = 1+t-2t^2$, so $f(t+h) = 1+(t+h)-2(t+h)^2$. From here, you need to evaluate
$$lim_{h to 0}frac{left[1+(t+h)-2(t+h)^2right]-left[1+t-2t^2right]}{h}$$
Simply note that $(a+b)^2 = a^2+2ab+b^2$ which you most probably know. You can solve this simply as the limit simplifies very quickly.
answered Jan 15 at 13:38
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
How would I prove that: d/dr (1+t-2t^2) = 1-4t
I assume you want to find the derivative with respect to $t$, not $r$.
Using differentiation from first principles.
I tried to integrate the equation and got the following:
f(t) =(1t+.5t^2-2/3t^3)
Why would you integrate if you want to differentiate (from first principles or otherwise)...?
Then I tried to uses the equation:
f(t+h)-f(t) / h
That's better, use the definition and find the following limit:
$$lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
for $f(t) = 1+t-2t^2$.
Is this correct and what do I do after this.
Use $f$ to evaluate $f(t+h)$ and $f(t)$ in the limit above: substitute and simplify first.
answered Jan 15 at 13:13
StackTDStackTD
22.9k2151
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
Yes, that’s the First Principle, or the definition of a derivative:
$$f’(t) = lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
You have $f(t) = 1+t-2t^2$, so $f(t+h) = 1+(t+h)-2(t+h)^2$. From here, you need to evaluate
$$lim_{h to 0}frac{left[1+(t+h)-2(t+h)^2right]-left[1+t-2t^2right]}{h}$$
Simply note that $(a+b)^2 = a^2+2ab+b^2$ which you most probably know. You can solve this simply as the limit simplifies very quickly.
answered Jan 15 at 13:38
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
Yes, that’s the First Principle, or the definition of a derivative:
$$f’(t) = lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
You have $f(t) = 1+t-2t^2$, so $f(t+h) = 1+(t+h)-2(t+h)^2$. From here, you need to evaluate
$$lim_{h to 0}frac{left[1+(t+h)-2(t+h)^2right]-left[1+t-2t^2right]}{h}$$
Simply note that $(a+b)^2 = a^2+2ab+b^2$ which you most probably know. You can solve this simply as the limit simplifies very quickly.
answered Jan 15 at 13:38
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
Yes, that’s the First Principle, or the definition of a derivative:
$$f’(t) = lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
You have $f(t) = 1+t-2t^2$, so $f(t+h) = 1+(t+h)-2(t+h)^2$. From here, you need to evaluate
$$lim_{h to 0}frac{left[1+(t+h)-2(t+h)^2right]-left[1+t-2t^2right]}{h}$$
Simply note that $(a+b)^2 = a^2+2ab+b^2$ which you most probably know. You can solve this simply as the limit simplifies very quickly.
answered Jan 15 at 13:38
KM101KM101
6,0251525
$endgroup$
Yes, that’s the First Principle, or the definition of a derivative:
$$f’(t) = lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
You have $f(t) = 1+t-2t^2$, so $f(t+h) = 1+(t+h)-2(t+h)^2$. From here, you need to evaluate
$$lim_{h to 0}frac{left[1+(t+h)-2(t+h)^2right]-left[1+t-2t^2right]}{h}$$
Simply note that $(a+b)^2 = a^2+2ab+b^2$ which you most probably know. You can solve this simply as the limit simplifies very quickly.
answered Jan 15 at 13:38
KM101KM101
6,0251525
answered Jan 15 at 13:38
KM101KM101
6,0251525
answered Jan 15 at 13:38
KM101KM101
6,0251525
answered Jan 15 at 13:38
KM101KM101
6,0251525
6,0251525
add a comment |
add a comment |
$begingroup$
How would I prove that: d/dr (1+t-2t^2) = 1-4t
I assume you want to find the derivative with respect to $t$, not $r$.
Using differentiation from first principles.
I tried to integrate the equation and got the following:
f(t) =(1t+.5t^2-2/3t^3)
Why would you integrate if you want to differentiate (from first principles or otherwise)...?
Then I tried to uses the equation:
f(t+h)-f(t) / h
That's better, use the definition and find the following limit:
$$lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
for $f(t) = 1+t-2t^2$.
Is this correct and what do I do after this.
Use $f$ to evaluate $f(t+h)$ and $f(t)$ in the limit above: substitute and simplify first.
answered Jan 15 at 13:13
StackTDStackTD
22.9k2151
$endgroup$
add a comment |
$begingroup$
How would I prove that: d/dr (1+t-2t^2) = 1-4t
I assume you want to find the derivative with respect to $t$, not $r$.
Using differentiation from first principles.
I tried to integrate the equation and got the following:
f(t) =(1t+.5t^2-2/3t^3)
Why would you integrate if you want to differentiate (from first principles or otherwise)...?
Then I tried to uses the equation:
f(t+h)-f(t) / h
That's better, use the definition and find the following limit:
$$lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
for $f(t) = 1+t-2t^2$.
Is this correct and what do I do after this.
Use $f$ to evaluate $f(t+h)$ and $f(t)$ in the limit above: substitute and simplify first.
answered Jan 15 at 13:13
StackTDStackTD
22.9k2151
$endgroup$
add a comment |
$begingroup$
How would I prove that: d/dr (1+t-2t^2) = 1-4t
I assume you want to find the derivative with respect to $t$, not $r$.
Using differentiation from first principles.
I tried to integrate the equation and got the following:
f(t) =(1t+.5t^2-2/3t^3)
Why would you integrate if you want to differentiate (from first principles or otherwise)...?
Then I tried to uses the equation:
f(t+h)-f(t) / h
That's better, use the definition and find the following limit:
$$lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
for $f(t) = 1+t-2t^2$.
Is this correct and what do I do after this.
Use $f$ to evaluate $f(t+h)$ and $f(t)$ in the limit above: substitute and simplify first.
answered Jan 15 at 13:13
StackTDStackTD
22.9k2151
$endgroup$
How would I prove that: d/dr (1+t-2t^2) = 1-4t
I assume you want to find the derivative with respect to $t$, not $r$.
Using differentiation from first principles.
I tried to integrate the equation and got the following:
f(t) =(1t+.5t^2-2/3t^3)
Why would you integrate if you want to differentiate (from first principles or otherwise)...?
Then I tried to uses the equation:
f(t+h)-f(t) / h
That's better, use the definition and find the following limit:
$$lim_{h to 0}frac{f(t+h)-f(t)}{h}$$
for $f(t) = 1+t-2t^2$.
Is this correct and what do I do after this.
Use $f$ to evaluate $f(t+h)$ and $f(t)$ in the limit above: substitute and simplify first.
answered Jan 15 at 13:13
StackTDStackTD
22.9k2151
answered Jan 15 at 13:13
StackTDStackTD
22.9k2151
answered Jan 15 at 13:13
StackTDStackTD
22.9k2151
answered Jan 15 at 13:13
StackTDStackTD
22.9k2151
22.9k2151
add a comment |
add a comment |
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