Find f'(x): f'(x) = ##\frac {dy} {dx},\frac {dx}{dt}= \frac {8t^7}{1+t^{16}} ##

[Mark44]

you are so quick in telling me am wrong

Because you have written so many incorrect things in this thread.
Here are three of them:

$F'(x)$= $\ {d} /{dt} \int_ a^x f(t) \ dt$= $f(x)$

The integral $\int_ a^x f(t) \ dt$ is a function of x, not t, so it doesn't make sense to differentiate with respect to t. Also, without defining what F and F' represent, it doesn't make sense to start an equation off with either of them.

then $f'(x) = f(t)$

As already noted by another responder, this doesn't make sense.

let $u = 1+t^{16}$

→ $\int_x^3 0.5 u^{-0.5}du$

As I said earlier, this isn't how substitution works. It's not enough to just define u, you have to also find du, which in this case is $16t^{15}dt$. Since the original integral doesn't have $16t^{15}$, substitution won't work in this case.

i last did this thing 10years ago, just need refreshing my brain

Like I said, it would be a good idea to review integration by substitution. You may have been proficient in this technique 10 years ago, but you're pretty rusty with it now.

i have no problem absolutely with using integration by substitution

Any other approach? on this problem?

It's possible a trig substitution would work, but using the Fund. Theorem is vastly simpler here.

 

[Mark44]

have you checked...

$\int_a^b {f{x}}dx$ = - $\int_b^a {f{x}} dx$??

What are you asking? This is what I was talking about toward the end of post #23, about changing the order of the limits of integration.
Correction to the above: $\int_a^b f(x)dx$ = - $\int_b^a f(x) dx$

you can use this in the same problem as it is , changing the limits the negatives cancel out...Can you see?
...

In terms of your problem, $f(x) = \int_x^3 \sqrt{1 + t^{16}} dt = - \int_3^x \sqrt{1 + t^{16}} dt$
$\Rightarrow f'(x) = \frac d {dx}\left( - \int_3^x \sqrt{1 + t^{16}} dt \right) = -\sqrt{1 + x^{16}}$

 

[chwala]

What are you asking? This is what I was talking about toward the end of post #23, about changing the order of the limits of integration.
Correction to the above: $\int_a^b f(x)dx$ = - $\int_b^a f(x) dx$

In terms of your problem, $f(x) = \int_x^3 \sqrt{1 + t^{16}} dt = - \int_3^x \sqrt{1 + t^{16}} dt$
$\Rightarrow f'(x) = \frac d {dx}\left( - \int_3^x \sqrt{1 + t^{16}} dt \right) = -\sqrt{1 + x^{16}}$

Are you by any chance implying that we get the same result even after interchanging or switching the limits? Have you looked at the Leibniz integral (on the fundamental laws of calculus)? as indicated in post 9..looks like you don't know it, i am not formulating the chain rule in any way, rather i am using the theorem as it is. Kindly clarify, thanks Mark.

 

[chwala]

i have no problem absolutely with using integration by substitution. The problem here does not need it anyway. I made a few mistakes, i am human that's all

I used lebniz integral on fundamental laws of calculus here, look at the theorem kindly before responding..

 

[chwala]

Mark have a look below on attachment on leibniz integral or is the theorem wrong?

 

[chwala]

Mark have a look below on attachment on leibniz integral or is the theorem wrong?

in reference to your post 29.

 

[Ray Vickson]

Mark have a look below on attachment on leibniz integral or is the theorem wrong?

If you want people to look at attachments you should first convert them to jpeg pictures or pdf files. I for one, will never open a word attachment in this forum. Better still, just use LaTeX and type it all out yourself.

BTW: of course every helper here knows Leibnitz theorem---most having spent years teaching the material----but you do not show how the theorem is relevant to your problem. What are g(x) and h(x) in this context? Do not make us guess what you are trying to do.

 

[Mark44]

I used lebniz integral on fundamental laws of calculus here, look at the theorem kindly before responding..

Here is from what you wrote in post #24:

$f'(x)= f(h(x).h'(x)) - f(g(x).g'(x))$

First, whether you know it or not, each term above is using the chain rule, but is not written correctly, due to misplaced parentheses.
In the first term on the right, you are multiplying h(x) and h'(x), and then taking that value as the input to f. In the second term you are doing something similar.
This is what you wrote for the first term above: $f\left(h(x) \cdot h'(x)\right)$.
It should instead be $f(h(x)) \cdot h'(x)$.
Second, by using the chain rule, you are overcomplicating things. As @Ray Vickson notes, you wrote the expression above using g(x) and h(x) without explaining what they were. The limits of integration in the original problem are simply x and 3, so there is no need to complicate matters by bringing in extra functions g(x) and h(x), and hence, no need for the chain rule.

Are you by any chance implying that we get the same result even after interchanging or switching the limits?

?
If the limits of integration are switched, you get a value whose sign is the opposite. That is clearly shown in post #32.

 

[chwala]

If you want people to look at attachments you should first convert them to jpeg pictures or pdf files. I for one, will never open a word attachment in this forum. Better still, just use LaTeX and type it all out yourself.

BTW: of course every helper here knows Leibnitz theorem---most having spent years teaching the material----but you do not show how the theorem is relevant to your problem. What are g(x) and h(x) in this context? Do not make us guess what you are trying to do.

good in my approach, $h(x) = 3 and g(x) = x$ this is what i used and i got a negative answer, i however changed the limits to get the required or correct value using post 30...what am simply saying with due respect don't be so quick in general, saying i am wrong without understanding my post...again i am not used to your language latex...its like me requiring you to type your work in my african language of which i know you will not understand, am trying atleast...

 

[chwala]

Here is from what you wrote in post #24:

First, whether you know it or not, each term above is using the chain rule, but is not written correctly, due to misplaced parentheses.
In the first term on the right, you are multiplying h(x) and h'(x), and then taking that value as the input to f. In the second term you are doing something similar.
This is what you wrote for the first term above: $f\left(h(x) \cdot h'(x)\right)$.
It should instead be $f(h(x)) \cdot h'(x)$.
Second, by using the chain rule, you are overcomplicating things. As @Ray Vickson notes, you wrote the expression above using g(x) and h(x) without explaining what they were. The limits of integration in the original problem are simplly x and 3, so there is no need to complicate matters by bringing in extra functions g(x) and h(x), and hence, no need for the chain rule.

?
If the limits of integration are switched, you get a value whose sign is the opposite. That is clearly shown in post #32.

latex is your language, i don't fully understand it, a few errors here and there shouldn't distort my intention, can you understand my african language if i would require you to do so?

 

[chwala]

good in my approach, $h(x) = 3 and g(x) = x$ this is what i used and i got a negative answer, i however changed the limits to get the required or correct value using post 30...what am simply saying with due respect don't be so quick in general, saying i am wrong without understanding my post...again i am not used to your language latex...its like me requiring you to type your work in my african language of which i know you will not understand, am trying atleast...

or am i wrong to say $h(x) = 3$, 3 being a constant?

 

[Ray Vickson]

latex is your language, i don't fully understand it, a few errors here and there shouldn't distort my intention, can you understand my african language if i would require you to do so?

No, LaTeX is not a language; it is a typesetting system. People from all over the world use it, whether they are English, French, Algerian, Russian, Nigerian, whatever.

Anyway, using plain text is good enough if you use parentheses and some standard conventions. In plain text, Leibnitz says that
(d/dx) int_{t=g(x) .. h(x)} f(t) dt = h'(x) f(h(x)) - g'(x) f(g(x))

 

[chwala]

No, LaTeX is not a language; it is a typesetting system. People from all over the world use it, whether they are English, French, Algerian, Russian, Nigerian, whatever.

Anyway, using plain text is good enough if you use parentheses and some standard conventions. In plain text, Leibnitz says that
(d/dx) int_{t=g(x) .. h(x)} f(t) dt = h'(x) f(h(x)) - g'(x) f(g(x))

ok was my substitution correct in regards to the theorem that i used?

 

[chwala]

ok was my substitution correct in regards to the theorem that i used?

nice weekend folks going to take some african brew, enjoy bye

 

[Ray Vickson]

ok was my substitution correct in regards to the theorem that i used?

Please indicate the message number you are referring to. There are so many that are all woven together that I cannot tell which post you are citing.

 

[Mark44]

a few errors here and there shouldn't distort my intention

The language is mathematics, which is universal. LaTeX is merely a way to render mathematical expressions to make them look like those that appear in textbooks. Since you are relatively inexperienced at using LaTeX, we have a tutorial here: https://www.physicsforums.com/help/latexhelp/

"A few errors here and there..." made it impossible to understand what you were doing. I commend you for going back over calculus that you studied 10 years ago, but it appears to me that you have either forgotten a lot of it or it wasn't very clear to you in the first place. We all are human, and so are prone to making mistakes, but we should try to work harder to prevent them, and not just brush them off as "a few errors here and there."

 

[Ericpoke]

No, LaTeX is not a language; it is a typesetting system. People from all over the world use it, whether they are English, French, Algerian, Russian, Nigerian, whatever.

Anyway, using plain text is good enough if you use parentheses and some standard conventions. In plain text, Leibnitz says that
(d/dx) int_{t=g(x) .. h(x)} f(t) dt = h'(x) f(h(x)) - g'(x) f(g(x))

Agreed and this is what I love about Math. Spoken and written languages might be different but Math usually shares the same type setting pretty much everywhere. Great point.

 

[chwala]

The language is mathematics, which is universal. LaTeX is merely a way to render mathematical expressions to make them look like those that appear in textbooks. Since you are relatively inexperienced at using LaTeX, we have a tutorial here: https://www.physicsforums.com/help/latexhelp/

"A few errors here and there..." made it impossible to understand what you were doing. I commend you for going back over calculus that you studied 10 years ago, but it appears to me that you have either forgotten a lot of it or it wasn't very clear to you in the first place. We all are human, and so are prone to making mistakes, but we should try to work harder to prevent them, and not just brush them off as "a few errors here and there."

lol Mark you impress me, calculus i have forgotten a lot of it! that sounds tough..i am quite good in calculus,linear algebra,statistics and probability,operations research,numerical analysis , vector calculus etc so it means if i can't solve a problem then i generally don't know? let me ask you a question, do you know everything in Mathematics yourself? enjoy your day

 

[chwala]

lol Mark you impress me, calculus i have forgotten a lot of it! that sounds tough..i am quite good in calculus,linear algebra,statistics and probability,operations research,numerical analysis , vector calculus etc so it means if i can't solve a problem then i generally don't know? let me ask you a question, do you know everything in Mathematics yourself? enjoy your day

but generally i respect you, i am still growing in this field, i am a toddler in matters mathematics. atleast am trying and moving forward, just to excite you about the different areas i enjoy in calculus..implicit differentiation, integration of trigonometric fxns, derivatives of trig fxs and their derivations that includes double angles, integration by substitution where we use chain rule and integration by parts infact i find them easy...