Question

Calc 2 help please! Find the derivative of d/dx (integral) (top limit) 3 sq rt x (bottom limit) x/6 cos t dt.... I know how to solve this with standard constants in the integral but the variable/ sq rt has thrown me off.

Answer 1

Oh really? ^_^
Then I suggest you solve it first with what you call 'standard constraints' :P

Solve the coloured bit, as I ever so slightly tweaked it :D
\[\Large \frac{d}{dx}\color{blue}{\int\limits_a^b\cos(t)dt}\]

There... those are what you'd call standard constraints, right? XD

Answer 2

well with that its just gonna be the deriv of cos t dt so cos t..

Answer 3

Don't mind the derivative d/dx first.
Just solve the blue integral.

Answer 4

Or... you still weren't taught how to take integrals?

Answer 5

the example before this we solved had the integral as t^9 upper 0 lower, and 3 sq rt u du to the right, so we simply reworked that to the derivative. so i'm confused on the integral part yeah

Answer 6

Okay, allow me to present you a formula then ^_^
\[\Large \frac{d}{dx} \int \limits_{a(x)}^{b(x)}f(t) dt = f\left(b(x)\right)b'(x) -f\left(a(x)\right)a'(x)\]

Answer 7

so input each limit on the integral times their derivative and take the difference?

Answer 8

Yup.
This is a result of the chain rule.
But I suppose you'll get to that later.
Before you do this, what was the answer to that other integral, the one with a t^9 upper limit?

Answer 9

9t^11

Answer 10

Are you sure it's 3sqrt?
Don't you mean cube root?

Answer 11

we did 3 sq rt u du -> u^ 1/3 du -> 3/4 u ^ 4/3 -> 3/4 (t^9) ^ 4/3 -> 3/4 t ^ 12 take the deriv of that and 9t^11

Answer 12

Like this:
\[\Large \sqrt[3]{\color{white}{\qquad}}\]

Answer 13

yeah thats what i meant sorry

Answer 14

so given we just have cos t dt i didnt know where to begin

Answer 15

Okay, well let me just demonstrate to you that the formula I gave you works (and is actually more general)
\[\Large \frac{d}{dt} \int \limits_{a(t)}^{b(t)}f(u) du = f\left(b(t)\right)b'(t) -f\left(a(t)\right)a'(t)\]

^this is the formula, tweaked in terms of u and t, right? :)

Answer 16

Okay, in this case, our a(t) is 0
b(t) is t^9
and \[\large f(u) = \sqrt[3]u \] correct?

Answer 17

yes.

Answer 18

Johnny...!
<snaps fingers>

Focus!

LOL
I'm TJ by the way ^_^

Answer 19

Oh, you're here... whoops :3

Answer 20

Now, \[\Large f\left(b(t)\right)= f(t^9) = \sqrt[3]{t^9} = t^3\]

Still with me?

Answer 21

why?
I just substituted...

Answer 22

meant yes

Answer 23

Right.
Now, b'(t) is simply the derivative of t^9 and this is just 9t^8

Right?

So we can replace this part:
\[\Large \frac{d}{dt} \int \limits_{0}^{t^9}\sqrt[3]u du =\color{red}{ f\left(b(t)\right)b'(t)} -f\left(a(t)\right)a'(t)\]
With this:
\[\Large \frac{d}{dt} \int \limits_{0}^{t^9}\sqrt[3]u du =\color{blue}{ (t^3)(9t^8)} -f\left(a(t)\right)a'(t)\]

Catch me so far?

Answer 24

roger

Answer 25

now, a(t) is simply 0
so actually, both f(a(t)) and a'(t) are going to be zero.
\[\Large \frac{d}{dt} \int \limits_{0}^{t^9}\sqrt[3]u du =\color{}{ (t^3)(9t^8)} -\cancel{f\left(a(t)\right)a'(t)}^0\]
So we have:
\[\Large \frac{d}{dt} \int \limits_{0}^{t^9}\sqrt[3]u du =\color{}{ (t^3)(9t^8)}=9t^{11}\]
And voila!
It worked ^_^

Answer 26

so for this question, cube root x is the upper, but we cant simplify that down, or would you have f(x^1/3)(x/3) for the first part??

Answer 27

f(b(x))b'(x)= (x^1/3)(x/3)?

Answer 28

Unfortunately :D
And slow down, lol
what's the derivative of the cube root of x?
I don't think it's it's as simple as x/3
:P

Answer 29

crap, 1/3x^2/3?

Answer 30

\[\Large \frac{d}{dx}x^\frac13 = \frac13x^{-\frac23}\]
:P

Answer 31

really? wheres the negative from?

Answer 32

Seriously? lol
\[\Large \frac{d}{dx}x^a = ax^{a-1}\]
And take \[\large a = \frac13\]
XD

Answer 33

ugh ok

Answer 34

Convinced?
Power rule, mate ^_^

Answer 35

so (x^1/3)(1/3x^-2/3) -

Answer 36

(x/6)(1/6)?

Answer 37

Nope :P

Answer 38

sigh

Answer 39

Remember, it's f(b(t))
not simply b(t) :D

Answer 40

In your case, what was f?
(f was the function being integrated, look on it now :) )
\[\Large \frac{d}{dx} \int \limits_{a(x)}^{b(x)}f(t) dt = f\left(b(x)\right)b'(x) -f\left(a(x)\right)a'(x)\]

Answer 41

so everything input should be multiplied by cos t?

Answer 42

not multiplied, silly.
evaluated under cosine XD

Answer 43

but how the heck can we eval 1/3x^ -2/3 under cos?

Answer 44

please study this formula more CAREFULLY
\[\Large \frac{d}{dx} \int \limits_{a(x)}^{b(x)}f(t) dt = f\left(b(x)\right)b'(x) -f\left(a(x)\right)a'(x)\]

What are evaluated under the f function?
Are they the derivatives of a and b?
NO!
Just a and b themselves.
And don't worry about evaluating, just express as is.

Answer 45

If you don't like it this way, then maybe this is clearer:
\[\Large \frac{d}{dx} \int \limits_{a(x)}^{b(x)}f(t) dt =[f\circ b](x)b'(x) -[f\circ a](x)a'(x)\]

Answer 46

lets stick to what we had, i think i just need to see the eventual simplification of it

Answer 47

Right... so what did you get, given my corrections...?

Answer 48

And leave the stuff inside the cosine as is, you can't possibly evaluate them to exactness...

Answer 49

so rather than x^1/3 * 1/3x^ -2/3 i should have that integrated with the cos t we started with?

Answer 50

(cos^1/3)(1/3cos^-2/3) such as that?

Answer 51

*cos x^1/3
Otherwise, yes. :)

Answer 52

ok so only the first part we are using the function got it sorry

Answer 53

So do you have a new answer? :)

Answer 54

(cos x^1/3)(1/3 cos^-2/3)-

Answer 55

(cos x/6)(1/6)

Answer 56

Bingo ^_^

You can arrive at that answer another way, however, if all you had trouble with was integrating cos, then just recall that
\[\large \int \cos(t) dt = \sin(t)+C\]
And evaluate at the limits... and then differentiate :D

Answer 57

roger

Answer 58

now for the simplification of that....

Answer 59

So basically the way to approach it was:
\[\Large \frac{d}{dx} \int\limits_\frac x6 ^ \sqrt[3]x \cos(t) dt = \frac{d}{dx}\left[\sin(\sqrt[3]x ) -\sin\left(\frac x6\right)\right]\]

Answer 60

As for simplification, it's only a few adjustments to the fractions, nothing major, and there can be no more simplification unless you know what x is.

Answer 61

they have the answer as 1/3x^-2/3 cos cube root of x,...so the front half of the formula we just did

Answer 62

Just the front half, eh?
That's really interesting... what could have caused the other 'half' to zero out, I wonder...

Are you sure you told us the right limits?

Answer 63

cube root of x upper, lower = pi/6 , wrote x /6 down in notes for some reason

Answer 64

which makes it cos pi/6 * 0 which cancels out yes, thank you though

Answer 65

Okay, then everything's settled :P