Finding the derivative of $x^sqrt{5}$ [closed]
$begingroup$
How could we find the first derivative of $$x^{sqrt5}$$ using the definition of the derivative? It is not allowed to use the rules.
I am a high school student and we are only allowed to answer the question using this definition of the derivative
$$ f'(x)=lim_{hto 0} frac{f(x+h)-f(x)}{h}$$
I stopped here :
$$lim_{hto 0} frac{(x+h)^{sqrt5}-x^{sqrt5}}{h}$$
calculus derivatives
edited Dec 8 '18 at 21:00
gimusi
92.9k94494
asked Dec 8 '18 at 19:40
H.MathH.Math
162
$endgroup$
closed as off-topic by Nosrati, Jyrki Lahtonen, Shaun, Trevor Gunn, RRL Dec 8 '18 at 23:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Jyrki Lahtonen, Shaun, Trevor Gunn, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 6 more comments
$begingroup$
How could we find the first derivative of $$x^{sqrt5}$$ using the definition of the derivative? It is not allowed to use the rules.
I am a high school student and we are only allowed to answer the question using this definition of the derivative
$$ f'(x)=lim_{hto 0} frac{f(x+h)-f(x)}{h}$$
I stopped here :
$$lim_{hto 0} frac{(x+h)^{sqrt5}-x^{sqrt5}}{h}$$
calculus derivatives
edited Dec 8 '18 at 21:00
gimusi
92.9k94494
asked Dec 8 '18 at 19:40
H.MathH.Math
162
$endgroup$
closed as off-topic by Nosrati, Jyrki Lahtonen, Shaun, Trevor Gunn, RRL Dec 8 '18 at 23:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Jyrki Lahtonen, Shaun, Trevor Gunn, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Do study the guide for new askers! Just making sure you have seen it :)
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 19:43
$begingroup$
Taht's really basic, hint will help you but in general you should add more detail about your doubts and effort on that. More interesting to solve later $(sqrt 5)^x$.
$endgroup$
– gimusi
Dec 8 '18 at 19:48
$begingroup$
To have some hint you should add some detail more about what you have tried.
$endgroup$
– gimusi
Dec 8 '18 at 20:04
2
$begingroup$
@H.Math I think that your teacher gave you a wrong assignement, it seems too advanced to me for high school level without using the exponential trick!
$endgroup$
– gimusi
Dec 8 '18 at 21:14
1
$begingroup$
+1 for the comment from @gimusi . We are curious about how your teacher will solve this problem for you when the class is stuck. Please come back here and post that answer when you get it.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 22:59
|
show 6 more comments
$begingroup$
How could we find the first derivative of $$x^{sqrt5}$$ using the definition of the derivative? It is not allowed to use the rules.
I am a high school student and we are only allowed to answer the question using this definition of the derivative
$$ f'(x)=lim_{hto 0} frac{f(x+h)-f(x)}{h}$$
I stopped here :
$$lim_{hto 0} frac{(x+h)^{sqrt5}-x^{sqrt5}}{h}$$
calculus derivatives
edited Dec 8 '18 at 21:00
gimusi
92.9k94494
asked Dec 8 '18 at 19:40
H.MathH.Math
162
$endgroup$
How could we find the first derivative of $$x^{sqrt5}$$ using the definition of the derivative? It is not allowed to use the rules.
I am a high school student and we are only allowed to answer the question using this definition of the derivative
$$ f'(x)=lim_{hto 0} frac{f(x+h)-f(x)}{h}$$
I stopped here :
$$lim_{hto 0} frac{(x+h)^{sqrt5}-x^{sqrt5}}{h}$$
calculus derivatives
calculus derivatives
edited Dec 8 '18 at 21:00
gimusi
92.9k94494
asked Dec 8 '18 at 19:40
H.MathH.Math
162
edited Dec 8 '18 at 21:00
gimusi
92.9k94494
asked Dec 8 '18 at 19:40
H.MathH.Math
162
edited Dec 8 '18 at 21:00
gimusi
92.9k94494
edited Dec 8 '18 at 21:00
gimusi
92.9k94494
edited Dec 8 '18 at 21:00
gimusi
92.9k94494
92.9k94494
asked Dec 8 '18 at 19:40
H.MathH.Math
162
asked Dec 8 '18 at 19:40
H.MathH.Math
162
asked Dec 8 '18 at 19:40
H.MathH.Math
162
162
closed as off-topic by Nosrati, Jyrki Lahtonen, Shaun, Trevor Gunn, RRL Dec 8 '18 at 23:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Jyrki Lahtonen, Shaun, Trevor Gunn, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Nosrati, Jyrki Lahtonen, Shaun, Trevor Gunn, RRL Dec 8 '18 at 23:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Jyrki Lahtonen, Shaun, Trevor Gunn, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Do study the guide for new askers! Just making sure you have seen it :)
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 19:43
$begingroup$
Taht's really basic, hint will help you but in general you should add more detail about your doubts and effort on that. More interesting to solve later $(sqrt 5)^x$.
$endgroup$
– gimusi
Dec 8 '18 at 19:48
$begingroup$
To have some hint you should add some detail more about what you have tried.
$endgroup$
– gimusi
Dec 8 '18 at 20:04
2
$begingroup$
@H.Math I think that your teacher gave you a wrong assignement, it seems too advanced to me for high school level without using the exponential trick!
$endgroup$
– gimusi
Dec 8 '18 at 21:14
1
$begingroup$
+1 for the comment from @gimusi . We are curious about how your teacher will solve this problem for you when the class is stuck. Please come back here and post that answer when you get it.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 22:59
|
show 6 more comments
3
$begingroup$
Do study the guide for new askers! Just making sure you have seen it :)
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 19:43
$begingroup$
Taht's really basic, hint will help you but in general you should add more detail about your doubts and effort on that. More interesting to solve later $(sqrt 5)^x$.
$endgroup$
– gimusi
Dec 8 '18 at 19:48
$begingroup$
To have some hint you should add some detail more about what you have tried.
$endgroup$
– gimusi
Dec 8 '18 at 20:04
2
$begingroup$
@H.Math I think that your teacher gave you a wrong assignement, it seems too advanced to me for high school level without using the exponential trick!
$endgroup$
– gimusi
Dec 8 '18 at 21:14
1
$begingroup$
+1 for the comment from @gimusi . We are curious about how your teacher will solve this problem for you when the class is stuck. Please come back here and post that answer when you get it.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 22:59
3
3
$begingroup$
Do study the guide for new askers! Just making sure you have seen it :)
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 19:43
$begingroup$
Do study the guide for new askers! Just making sure you have seen it :)
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 19:43
$begingroup$
Taht's really basic, hint will help you but in general you should add more detail about your doubts and effort on that. More interesting to solve later $(sqrt 5)^x$.
$endgroup$
– gimusi
Dec 8 '18 at 19:48
$begingroup$
Taht's really basic, hint will help you but in general you should add more detail about your doubts and effort on that. More interesting to solve later $(sqrt 5)^x$.
$endgroup$
– gimusi
Dec 8 '18 at 19:48
$begingroup$
To have some hint you should add some detail more about what you have tried.
$endgroup$
– gimusi
Dec 8 '18 at 20:04
$begingroup$
To have some hint you should add some detail more about what you have tried.
$endgroup$
– gimusi
Dec 8 '18 at 20:04
2
2
$begingroup$
@H.Math I think that your teacher gave you a wrong assignement, it seems too advanced to me for high school level without using the exponential trick!
$endgroup$
– gimusi
Dec 8 '18 at 21:14
$begingroup$
@H.Math I think that your teacher gave you a wrong assignement, it seems too advanced to me for high school level without using the exponential trick!
$endgroup$
– gimusi
Dec 8 '18 at 21:14
1
1
$begingroup$
+1 for the comment from @gimusi . We are curious about how your teacher will solve this problem for you when the class is stuck. Please come back here and post that answer when you get it.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 22:59
$begingroup$
+1 for the comment from @gimusi . We are curious about how your teacher will solve this problem for you when the class is stuck. Please come back here and post that answer when you get it.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 22:59
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
First, we write for $xge 0$
$$begin{align}
frac{(x+h)^sqrt 5-x^sqrt5}{h}&=x^sqrt 5left(frac{left(1+frac hxright)^sqrt5-1}{h}right)end{align}tag1$$
Next, using the inequalities (for the exponential and logarithmic functions) that I developed in THIS ANSWER using elementary analysis only, we note that for $x> 0$, $h<x/sqrt5$
$$frac{sqrt 5h}{x+h}le left(1+frac hxright)^sqrt5-1=e^{sqrt5 logleft(1+frac hxright)}-1le frac{sqrt 5h}{x-sqrt5h}tag2$$
Finally, using $(2)$ in $(1)$ and applying the squeeze theorem yields the coveted limit for $x>0$
$$lim_{hto0}frac{(x+h)^sqrt 5-x^sqrt5}{h}=sqrt5 x^{sqrt5-1}$$
as expected!
Tools Used:
$(i)$ $x^a=e^{alog(x)}$
$(ii)$ Inequalities obtained using the limit definition of the exponential function and Bernoulli's Inequality
$(iii)$ The Squeeze Theorem
edited Dec 8 '18 at 23:23
Trevor Gunn
14.3k32046
answered Dec 8 '18 at 22:50
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Very nice Mark....really the best thing we can do I think and you did it at best ;)
$endgroup$
– gimusi
Dec 9 '18 at 1:27
$begingroup$
I'm real curious to know whether the teacher of Hmath is looking for something like that! Bye
$endgroup$
– gimusi
Dec 9 '18 at 1:28
$begingroup$
@gimusi Thank you for the nice words. I don't really know the expectations of the OP's teacher. It's not likely that the approach herein is what that teacher had in mind. But it does provide a fairly elementary way forward. I don't know how to find the limit with a lower-level tool kit.
$endgroup$
– Mark Viola
Dec 9 '18 at 3:14
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First, we write for $xge 0$
$$begin{align}
frac{(x+h)^sqrt 5-x^sqrt5}{h}&=x^sqrt 5left(frac{left(1+frac hxright)^sqrt5-1}{h}right)end{align}tag1$$
Next, using the inequalities (for the exponential and logarithmic functions) that I developed in THIS ANSWER using elementary analysis only, we note that for $x> 0$, $h<x/sqrt5$
$$frac{sqrt 5h}{x+h}le left(1+frac hxright)^sqrt5-1=e^{sqrt5 logleft(1+frac hxright)}-1le frac{sqrt 5h}{x-sqrt5h}tag2$$
Finally, using $(2)$ in $(1)$ and applying the squeeze theorem yields the coveted limit for $x>0$
$$lim_{hto0}frac{(x+h)^sqrt 5-x^sqrt5}{h}=sqrt5 x^{sqrt5-1}$$
as expected!
Tools Used:
$(i)$ $x^a=e^{alog(x)}$
$(ii)$ Inequalities obtained using the limit definition of the exponential function and Bernoulli's Inequality
$(iii)$ The Squeeze Theorem
edited Dec 8 '18 at 23:23
Trevor Gunn
14.3k32046
answered Dec 8 '18 at 22:50
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Very nice Mark....really the best thing we can do I think and you did it at best ;)
$endgroup$
– gimusi
Dec 9 '18 at 1:27
$begingroup$
I'm real curious to know whether the teacher of Hmath is looking for something like that! Bye
$endgroup$
– gimusi
Dec 9 '18 at 1:28
$begingroup$
@gimusi Thank you for the nice words. I don't really know the expectations of the OP's teacher. It's not likely that the approach herein is what that teacher had in mind. But it does provide a fairly elementary way forward. I don't know how to find the limit with a lower-level tool kit.
$endgroup$
– Mark Viola
Dec 9 '18 at 3:14
add a comment |
$begingroup$
First, we write for $xge 0$
$$begin{align}
frac{(x+h)^sqrt 5-x^sqrt5}{h}&=x^sqrt 5left(frac{left(1+frac hxright)^sqrt5-1}{h}right)end{align}tag1$$
Next, using the inequalities (for the exponential and logarithmic functions) that I developed in THIS ANSWER using elementary analysis only, we note that for $x> 0$, $h<x/sqrt5$
$$frac{sqrt 5h}{x+h}le left(1+frac hxright)^sqrt5-1=e^{sqrt5 logleft(1+frac hxright)}-1le frac{sqrt 5h}{x-sqrt5h}tag2$$
Finally, using $(2)$ in $(1)$ and applying the squeeze theorem yields the coveted limit for $x>0$
$$lim_{hto0}frac{(x+h)^sqrt 5-x^sqrt5}{h}=sqrt5 x^{sqrt5-1}$$
as expected!
Tools Used:
$(i)$ $x^a=e^{alog(x)}$
$(ii)$ Inequalities obtained using the limit definition of the exponential function and Bernoulli's Inequality
$(iii)$ The Squeeze Theorem
edited Dec 8 '18 at 23:23
Trevor Gunn
14.3k32046
answered Dec 8 '18 at 22:50
Mark ViolaMark Viola
131k1275171
$endgroup$
$begingroup$
Very nice Mark....really the best thing we can do I think and you did it at best ;)
$endgroup$
– gimusi
Dec 9 '18 at 1:27
$begingroup$
I'm real curious to know whether the teacher of Hmath is looking for something like that! Bye
$endgroup$
– gimusi
Dec 9 '18 at 1:28
$begingroup$
@gimusi Thank you for the nice words. I don't really know the expectations of the OP's teacher. It's not likely that the approach herein is what that teacher had in mind. But it does provide a fairly elementary way forward. I don't know how to find the limit with a lower-level tool kit.
$endgroup$
– Mark Viola
Dec 9 '18 at 3:14
add a comment |
$begingroup$
First, we write for $xge 0$
$$begin{align}
frac{(x+h)^sqrt 5-x^sqrt5}{h}&=x^sqrt 5left(frac{left(1+frac hxright)^sqrt5-1}{h}right)end{align}tag1$$
Next, using the inequalities (for the exponential and logarithmic functions) that I developed in THIS ANSWER using elementary analysis only, we note that for $x> 0$, $h<x/sqrt5$
$$frac{sqrt 5h}{x+h}le left(1+frac hxright)^sqrt5-1=e^{sqrt5 logleft(1+frac hxright)}-1le frac{sqrt 5h}{x-sqrt5h}tag2$$
Finally, using $(2)$ in $(1)$ and applying the squeeze theorem yields the coveted limit for $x>0$
$$lim_{hto0}frac{(x+h)^sqrt 5-x^sqrt5}{h}=sqrt5 x^{sqrt5-1}$$
as expected!
Tools Used:
$(i)$ $x^a=e^{alog(x)}$
$(ii)$ Inequalities obtained using the limit definition of the exponential function and Bernoulli's Inequality
$(iii)$ The Squeeze Theorem
edited Dec 8 '18 at 23:23
Trevor Gunn
14.3k32046
answered Dec 8 '18 at 22:50
Mark ViolaMark Viola
131k1275171
$endgroup$
First, we write for $xge 0$
$$begin{align}
frac{(x+h)^sqrt 5-x^sqrt5}{h}&=x^sqrt 5left(frac{left(1+frac hxright)^sqrt5-1}{h}right)end{align}tag1$$
Next, using the inequalities (for the exponential and logarithmic functions) that I developed in THIS ANSWER using elementary analysis only, we note that for $x> 0$, $h<x/sqrt5$
$$frac{sqrt 5h}{x+h}le left(1+frac hxright)^sqrt5-1=e^{sqrt5 logleft(1+frac hxright)}-1le frac{sqrt 5h}{x-sqrt5h}tag2$$
Finally, using $(2)$ in $(1)$ and applying the squeeze theorem yields the coveted limit for $x>0$
$$lim_{hto0}frac{(x+h)^sqrt 5-x^sqrt5}{h}=sqrt5 x^{sqrt5-1}$$
as expected!
Tools Used:
$(i)$ $x^a=e^{alog(x)}$
$(ii)$ Inequalities obtained using the limit definition of the exponential function and Bernoulli's Inequality
$(iii)$ The Squeeze Theorem
edited Dec 8 '18 at 23:23
Trevor Gunn
14.3k32046
answered Dec 8 '18 at 22:50
Mark ViolaMark Viola
131k1275171
edited Dec 8 '18 at 23:23
Trevor Gunn
14.3k32046
edited Dec 8 '18 at 23:23
Trevor Gunn
14.3k32046
edited Dec 8 '18 at 23:23
Trevor Gunn
14.3k32046
14.3k32046
answered Dec 8 '18 at 22:50
Mark ViolaMark Viola
131k1275171
answered Dec 8 '18 at 22:50
Mark ViolaMark Viola
131k1275171
answered Dec 8 '18 at 22:50
Mark ViolaMark Viola
131k1275171
131k1275171
$begingroup$
Very nice Mark....really the best thing we can do I think and you did it at best ;)
$endgroup$
– gimusi
Dec 9 '18 at 1:27
$begingroup$
I'm real curious to know whether the teacher of Hmath is looking for something like that! Bye
$endgroup$
– gimusi
Dec 9 '18 at 1:28
$begingroup$
@gimusi Thank you for the nice words. I don't really know the expectations of the OP's teacher. It's not likely that the approach herein is what that teacher had in mind. But it does provide a fairly elementary way forward. I don't know how to find the limit with a lower-level tool kit.
$endgroup$
– Mark Viola
Dec 9 '18 at 3:14
add a comment |
$begingroup$
Very nice Mark....really the best thing we can do I think and you did it at best ;)
$endgroup$
– gimusi
Dec 9 '18 at 1:27
$begingroup$
I'm real curious to know whether the teacher of Hmath is looking for something like that! Bye
$endgroup$
– gimusi
Dec 9 '18 at 1:28
$begingroup$
@gimusi Thank you for the nice words. I don't really know the expectations of the OP's teacher. It's not likely that the approach herein is what that teacher had in mind. But it does provide a fairly elementary way forward. I don't know how to find the limit with a lower-level tool kit.
$endgroup$
– Mark Viola
Dec 9 '18 at 3:14
$begingroup$
Very nice Mark....really the best thing we can do I think and you did it at best ;)
$endgroup$
– gimusi
Dec 9 '18 at 1:27
$begingroup$
Very nice Mark....really the best thing we can do I think and you did it at best ;)
$endgroup$
– gimusi
Dec 9 '18 at 1:27
$begingroup$
I'm real curious to know whether the teacher of Hmath is looking for something like that! Bye
$endgroup$
– gimusi
Dec 9 '18 at 1:28
$begingroup$
I'm real curious to know whether the teacher of Hmath is looking for something like that! Bye
$endgroup$
– gimusi
Dec 9 '18 at 1:28
$begingroup$
@gimusi Thank you for the nice words. I don't really know the expectations of the OP's teacher. It's not likely that the approach herein is what that teacher had in mind. But it does provide a fairly elementary way forward. I don't know how to find the limit with a lower-level tool kit.
$endgroup$
– Mark Viola
Dec 9 '18 at 3:14
$begingroup$
@gimusi Thank you for the nice words. I don't really know the expectations of the OP's teacher. It's not likely that the approach herein is what that teacher had in mind. But it does provide a fairly elementary way forward. I don't know how to find the limit with a lower-level tool kit.
$endgroup$
– Mark Viola
Dec 9 '18 at 3:14
add a comment |
3
$begingroup$
Do study the guide for new askers! Just making sure you have seen it :)
$endgroup$
– Jyrki Lahtonen
Dec 8 '18 at 19:43
$begingroup$
Taht's really basic, hint will help you but in general you should add more detail about your doubts and effort on that. More interesting to solve later $(sqrt 5)^x$.
$endgroup$
– gimusi
Dec 8 '18 at 19:48
$begingroup$
To have some hint you should add some detail more about what you have tried.
$endgroup$
– gimusi
Dec 8 '18 at 20:04
2
$begingroup$
@H.Math I think that your teacher gave you a wrong assignement, it seems too advanced to me for high school level without using the exponential trick!
$endgroup$
– gimusi
Dec 8 '18 at 21:14
1
$begingroup$
+1 for the comment from @gimusi . We are curious about how your teacher will solve this problem for you when the class is stuck. Please come back here and post that answer when you get it.
$endgroup$
– Ethan Bolker
Dec 8 '18 at 22:59