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

[chwala]

Homework Statement

if $f(x)= \int_x^3 \sqrt {1+t^{16}} dt$ find f'(x)

Homework Equations

The Attempt at a Solution

let $x = 1+t^{16},$ → $\frac {dx}{dt},= 16t^{15},$
$y= x^{1/2}$ →$\frac {dy} {dx},= 0.5x^{-0.5}$

 

[Eclair_de_XII]

I don't think you're asked to find $\frac{dx}{dt}$ or $\frac{dy}{dx}$. You're asked to find $\frac{df(x)}{dx}=f'(x)$.

In any case, why not review the Fundamental Theorem of Calculus to try and solve this problem?

 

[chwala]

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

 

[Eclair_de_XII]

I don't think that's right. How can a function of $x$ be equal to a function of $t$; that is, if neither are constant? Anyway, that's beside the point. Why don't you start by stating the first part of the Fundamental Theorem of Calculus? I believe that that's mostly what's needed in this problem.

 

[Mark44]

Homework Statement

if $f(x)= \int_x^3 \sqrt {1+t^{16}} dt$ find f'(x)

Homework Equations

The Attempt at a Solution

let $x = 1+t^{16},$ → $\frac {dx}{dt},= 16t^{15},$
$y= x^{1/2}$ →$\frac {dy} {dx},= 0.5x^{-0.5}$

No, this is not the right approach. You should not attempt to evaluate this integral directly. You should be using the Fundamental Theorem of Calculus to find f'(x), as @Eclair_de_XII has suggested.

 

[Mark44]

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

The integral is actually a function of x, due to the fact that x appears as one of the limits of integration. "t" is a "dummy" variable.

 

[chwala]

ok $F(x)$=$\int_ a^x { f(t)} dt$
then
→$F'(X) = f(x)$ for all $x∈ (a,x)$

 

[ubergewehr273]

Have you tried using Newton Liebnitz rule ?

 

[Mark44]

ok $F(x)$=$\int_ a^x { f(t)} dt$
then
→$F'(X) = f(x)$ for all $x∈ (a,x)$

No. The posted problem is this: if $f(x)= \int_x^3 \sqrt {1+t^{16}} dt$ find f'(x).
What is f'(x) in terms of the above?

 

[Mark44]

Have you tried using Newton Liebnitz rule ?

This is essentially the same advice that has already been given.

 

[chwala]

I don't know the rule, kindly show me I am stuck

 

[Eclair_de_XII]

Apply this part of the theorem that you posted:

$F'(x) = f(x)$ for all $x∈(a,x)$

to this part of the theorem that you also posted:

$F(x) =\int_a^xf(t)dt$

If you know how to do that, then you should be able to do the problem that you posted, given that you understand the relationship between $f(x)$ and $f'(x)$.

 

[chwala]

now $\frac {d} {dt} \int_x^3 \sqrt{1+t^{16}}$ dt
$= \int_x^3 0.5× 16t^{15}$/$\sqrt{1+t^{16}} dt$
let $u = 1+t^{16}$
→ $\int_x^3 0.5 u^{-0.5}du$
→$√1+t^{16}$ from x to 3
$= √4-√{1+x}$ = f(x)
f'(x)= $-0.5/ √{1+x}$

 

[chwala]

now $\frac {d} {dt} \int_x^3 √{1+t^{16}}$ dt
$= \int_x^3 0.5× 16t^{15}$/$√1+t^{16}$ dt
let $u = 1+t^{16}$
→ $\int_x^3 0.5 u^{-0.5}du$
→$√1+t^{16}$ from x to 3
$= √4-√{1+x}$ = f(x)
f'(x)= $-0.5/ √{1+x}$

note square root sign should cover all terms under it, sorry for using small square root sign..

 

[Mark44]

I don't know the rule, kindly show me I am stuck

Do you have a textbook? If so, look up "Fundamental Theorem of Calculus". If you can't find it in your textbook, look up this phrase on the internet.

now $\frac {d} {dt} \int_x^3 \sqrt{1+t^{16}}$ dt
$= \int_x^3 0.5× 16t^{15}$/$\sqrt{1+t^{16}} dt$

No. This is wrong.

let $u = 1+t^{16}$
→ $\int_x^3 0.5 u^{-0.5}du$

No. You aren't using the substitution method correctly. It isn't of any help in this problem.

→$√1+t^{16}$ from x to 3
$= √4-√{1+x}$ = f(x)
f'(x)= $-0.5/ √{1+x}$

 

[chwala]

where am i using the substitution wrongly...if i got the concept, its better you just pin point any error ..

 

[Eclair_de_XII]

I'd personally recommend just locating the Wikipedia article for it and looking for an explanation on how the theorem is used.

But I'm not sure if recommending that is kosher at this point.

 

[Mark44]

where am i using the substitution wrongly...if i got the concept, its better you just pin point any error ..

In post #14, you said
let $u = 1 + t^{16}$
You also have to find du, which you didn't do. You can't just replace dt with du here.
It seems that you don't have a good understanding of integration by substitution.

At any rate, integration by substitution won't work.

I'd personally recommend just locating the Wikipedia article for it and looking for an explanation on how it's used, but I'm not sure if recommending that is kosher at this point.

This is enough of a hint. Let @chwala do a bit of digging now.

 

[Eclair_de_XII]

This is enough of a hint.

Apologies. I've deleted my message.

 

[Mark44]

Apologies. I've deleted my message.

No need to apologize. Your post was OK. I undeleted the 2nd of the two posts you deleted.

 

[chwala]

ooooh i see...ok using the leibniz integral
$\frac {d} {dx},= \int_x^3 \sqrt {1+t^{16}}$ dt=
$\sqrt {1+3^{16}}$×16×${3^{15}}$-$\sqrt {1+x^{16}}$×1=
$1.506 ×10^{12}-\sqrt {1+x^{16}}$

 

[chwala]

ooooh i see...ok using the leibniz integral
$\frac {d} {dx},= \int_x^3 \sqrt {1+t^{16}}$ dt=
$\sqrt {1+3^{16}}$×16×${3^{15}}$-$\sqrt {1+x^{16}}$×1=
$1.506 ×10^{12}-\sqrt {1+x^{16}}$

is this correct?

 

[Mark44]

ooooh i see...ok using the leibniz integral
$\frac {d} {dx},= \int_x^3 \sqrt {1+t^{16}}$ dt=
$\sqrt {1+3^{16}}$×16×${3^{15}}$-$\sqrt {1+x^{16}}$×1=

The part just above is wrong.

$1.506 ×10^{12}-\sqrt {1+x^{16}}$

No, but this is closer than your previous attempts.

The Fundamental Theorem of Calculus says, in part, the following.
If $f(x) = \int_a^x g(t)dt$, then $f'(x) = g(x)$
Notice that a doesn't play a part in the answer. Also note the positions of the two integration limits, a and x. If they are in the opposite positions, you get a different answer.

In light of the mistakes you've made in this thread, it would be a very good idea to go back and review how integration by substitution works. Presumably that has been presented to you earlier, but it's clear you need some more practice with it. It's also good practice to check your work when you integrate, by differentiating your answer to an indefinite integral. When you differentiate your answer, you should get back to the integrand. If you don't, then you've made a mistake.

 

[chwala]

i made a mistake, the solution is correct only that it has a negative sign,
$f'(x)= f(h(x).h'(x)) - f(g(x).g'(x))$
this gives
$\sqrt {1+3^{16}}$.0 - $\sqrt {1+x^{16}}$.1
$= -\sqrt {1+x^{16}}$ which is correct the correct answer is $\sqrt {1+x^{16}}$
you are so quick in telling me am wrong, i last did this thing 10years ago, just need refreshing my brain...i will post the other method here, be patient

 

[chwala]

i made a mistake, the solution is correct only that it has a negative sign,
$f'(x)= f(h(x).h'(x)) - f(g(x).g'(x))$
this gives
$\sqrt {1+3^{16}}$.0 - $\sqrt {1+x^{16}}$.1
$= -\sqrt {1+x^{16}}$ which is correct the correct answer is $\sqrt {1+x^{16}}$
you are so quick in telling me am wrong, i last did this thing 10years ago, just need refreshing my brain...i will post the other method here, be patient

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

 

[chwala]

correct solution
$F'(x)$= $\ {d} /{dt} \int_ a^x f(t) \ dt$= $f(x)$
substituting directly by replacing $t$ with $x$,
$F'(x) = \sqrt {1+x^{16}}$ =$f(x)$

 

[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

aaaaah i can now see when you interchange the limits you get a positive value......

 

[chwala]

aaaaah i can now see when you interchange the limits you get a positive value......

Any other approach? on this problem?

 

[Mark44]

i made a mistake, the solution is correct only that it has a negative sign,
$f'(x)= f(h(x).h'(x)) - f(g(x).g'(x))$

I don't know what the above has to do with this problem. It's also not how the chain rule is formulated.

this gives
$\sqrt {1+3^{16}}$.0 - $\sqrt {1+x^{16}}$.1

No, the first term shouldn't be there

$= -\sqrt {1+x^{16}}$ which is correct the correct answer is $\sqrt {1+x^{16}}$
you are so quick in telling me am wrong, i last did this thing 10years ago, just need refreshing my brain...i will post the other method here, be patient

Here is what you wrote in post #1:
[/quote]if $f(x)= \int_x^3 \sqrt {1+t^{16}} dt$ find f'(x)

Notice that $\int_x^3 \sqrt {1+t^{16}} dt = -\int_3^x \sqrt {1+t^{16}} dt$, so this now matches the form of the Fundamental Theorem. If you miswrote the original integral, and it should be $\int_3^x \sqrt {1+t^{16}} dt$, then $\sqrt {1+x^{16}}$ is correct; otherwise, that negative sign should be there.

 

[chwala]

I don't know what the above has to do with this problem. It's also not how the chain rule is formulated.

No, the first term shouldn't be there

Here is what you wrote in post #1:

if $f(x)= \int_x^3 \sqrt {1+t^{16}} dt$ find f'(x)

Agreed Mark, the first constant term should not be there,otherwise the problem was written correctly but have you checked...

$\int_a^b {f{x}}dx$ = - $\int_b^a {f{x}} dx$?? you can use this in the same problem as it is , changing the limits the negatives cancel out...Can you see?

 

[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?