Need help

What don't you understand about the explanation? It seems rather straight forward to me.
 
JeffM said:
What don't you understand about the explanation? It seems rather straight forward to me.
Click to expand...
all I want to know is how to solve for the third differentiation "add a step and there substitute the value of u=2 in obtained second derivative of given function to get the required answer. Final answer: correct answer is 1/4 not 2/u^3" based on this feedback.

Here's my attempt
 

Attachments

  • Screenshot (64).png
    Screenshot (64).png
    17.6 KB · Views: 6
dollyayesha2345 said:
all I want to know is how to solve for the third differentiation "add a step and there substitute the value of u=2 in obtained second derivative of given function to get the required answer. Final answer: correct answer is 1/4 not 2/u^3" based on this feedback.
Click to expand...
[imath]{u^{ - 3}}\xrightarrow[{1st}]{D_x} - 3{u^{ - 4}}\xrightarrow[{2nd}]{D_x}12{u^{ - 5}}\xrightarrow[{3rd}]{D_x}-60{u^{ - 6}}[/imath]
 
Last edited:
No.

The third derivative is the derivative of the second derivative. In general, a derivative is a function rather than a specific number.

What is the second derivative according to the worked example? [math]y’’ = \dfrac{2}{u^3}.[/math]
[math]\text {Let } y‘’ = v(u) = 2u^{-3} \implies y’’’ = \dfrac{dv}{du} = (2)(-3)u^{(-3-1)} = -6u^{-4} = - \dfrac{6}{u^4}.[/math]
Now what is y’’’ when u = 2.
 
JeffM said:
No.

The third derivative is the derivative of the second derivative. In general, a derivative is a function rather than a specific number.

What is the second derivative according to the worked example? [math]y’’ = \dfrac{2}{u^3}.[/math]
[math]\text {Let } y‘’ = v(u) = 2u^{-3} \implies y’’’ = \dfrac{dv}{du} = (2)(-3)u^{(-3-1)} = -6u^{-4} = - \dfrac{6}{u^4}.[/math]
Now what is y’’’ when u = 2.
Click to expand...
Should I just solve and find all the three derivatives using u=2 and show?
(I am genuinely confused at this point)
What you have written above indeed is the way to find the third derivative, but it doesn't match the feedback I received. Now I am super confused to who is wrong(most probably its me).
? ?
 
Last edited:
dollyayesha2345 said:
Should I just solve and find all the three derivatives using u=2 and show?
(I am genuinely confused at this point)
What you have written above indeed is the way to find the third derivative, but it doesn't match the feedback I received. Now I am super confused to who is wrong(most probably its me).
? ?
Click to expand...
Dolly

It is always possible that I made a mistake. But you say you are not using a text or taking a course. You have not told us what the problem even is. You say that you have received feedback that disagrees with me.

Please give the complete problem exactly as it was given to you.

What was the feedback?

I cannot begin to straighten things out working in the dark.
 
JeffM said:
Dolly

It is always possible that I made a mistake. But you say you are not using a text or taking a course. You have not told us what the problem even is. You say that you have received feedback that disagrees with me.

Please give the complete problem exactly as it was given to you.

What was the feedback?

I cannot begin to straighten things out working in the dark.
Click to expand...
ok so I received a question which said "Compute [math]\frac{d^2y}{du^2}\ |_2[/math] for [math]y=\frac{1}{u}[/math]"
In comment #5 is my answer, where I was satisfied with finding 2 derivatives and submitted it. Then today morning I received a message saying that my answer is wrong reason being: "add a step and there substitute the value of u=2 in obtained second derivative of given function to get the required answer. Final answer: correct answer is 1/4 not 2/u^3". So I ended up starting this thread and has\dn't received any replies so I solved comment/reply#7 and ended up submitting it (didn't receive any feedback yet after my rework), so from you I have learned that it is wrong. Then you shared the correct third derivative but your answer doesn't match the feedback I received.

So now all I want to know is where should I substitute [math]u=2[/math] in order for my final answer to be [math]\frac{1}{4}[/math].?
 
Last edited:
dollyayesha2345 said:
ok so I received a question which said "Compute [math]\frac{d^2y}{du^2}\ |_2[/math] for [math]y=\frac{1}{u}[/math]"
In comment #5 is my answer, where I was satisfied with finding 2 derivatives and submitted it. Then today morning I received a message saying that my answer is wrong reason being: "add a step and there substitute the value of u=2 in obtained second derivative of given function to get the required answer. Final answer: correct answer is 1/4 not 2/u^3". So I ended up starting this thread and has\dn't received any replies so I solved comment/reply#7 and ended up submitting it (didn't receive any feedback yet after my rework), so from you I have learned that it is wrong. Then you shared the correct third derivative but your answer doesn't match the feedback I received.

So now all I want to know is where should I substitute [math]u=2[/math] in order for my final answer to be [math]\frac{1}{4}[/math].?
Click to expand...
The second derivative is [imath]\dfrac{2}{u^3}.[/imath]

So if u = 2, then the second derivative = [imath]\dfrac{2}{2^3} = \dfrac{2}{8} = \dfrac{1}{4}.[/imath]

And I did not give you a final answer for the third derivative, but left that for you to finish.
 
dollyayesha2345 said:
ok so I received a question which said "Compute [math]\frac{d^2y}{du^2}\ |_2[/math] for [math]y=\frac{1}{u}[/math]"
In comment #5 is my answer, where I was satisfied with finding 2 derivatives and submitted it. Then today morning I received a message saying that my answer is wrong reason being: "add a step and there substitute the value of u=2 in obtained second derivative of given function to get the required answer. Final answer: correct answer is 1/4 not 2/u^3". So I ended up starting this thread and has\dn't received any replies so I solved comment/reply#7 and ended up submitting it (didn't receive any feedback yet after my rework), so from you I have learned that it is wrong. Then you shared the correct third derivative but your answer doesn't match the feedback I received.

So now all I want to know is where should I substitute [math]u=2[/math] in order for my final answer to be [math]\frac{1}{4}[/math].?
Click to expand...
First you need to derive y" which is shorthand for \(\displaystyle \frac{d^2y}{dx^2}\) and then evaluate that at x = 2
 
JeffM said:
The second derivative is [imath]\dfrac{2}{u^3}.[/imath]

So if u = 2, then the second derivative = [imath]\dfrac{2}{2^3} = \dfrac{2}{8} = \dfrac{1}{4}.[/imath]

And I did not give you a final answer for the third derivative, but left that for you to finish.
Click to expand...
I'll genuinely be honest you have helped me a lot. Much much appreciated.
 
You must log in or register to reply here.
Top