Question

Please help! Calc! I do not understand this at all, and would greatly appreciate someone showing/explaining the steps: Find the derivative by A, evaluating the integral and differentiating the result, and B, by differentiating the integral directly. Problem: d/dt (integral) lower 0 upper t^9
^3 (sq rt) u * du

Answer 1

\(\large \displaystyle \frac{d}{dt} \int_{0}^{t^9} \sqrt[3]{u} \; \text{d} u \)
Is this what we are looking at?

Answer 2

Yes :D

Answer 3

Alright.
So the first method they ask you to take on is integrating it first and then taking the derivative.
So let's throw the d/dt to the wind and figure out how to integrate that integral first.
\( \large \displaystyle \color{#aaa}{\frac{d}{dt}} \int_{0}^{t^9} \sqrt[3]{u} \; \text{d} u \)

So,do you have any guesses as to how we would approach this integral?

Answer 4

find f(t), which I have as (integral) ^9 sq rt U du...but then my algebra skills are off or something cause my answer falls off

Answer 5

f(x)dx= f(b)-f(a) is what they are showing

Answer 6

Ah. That is what we do with the antiderivative once we know it.
But we have to find the antiderivative of the cube root of u.

Answer 7

I can simplify ^9 sq rt u du to u^1/9 du but then lost as to how they get to the next step

Answer 8

the example they have shows: ^5 sq rt u du -> u^1/5 du -> 5/6u^6/5 + C

Answer 9

that last step is where im lost

Answer 10

Getting from the integral of u^(1/5) du to 5/6 u^(6/5) ?

Answer 11

yes

Answer 12

It is the application of power rule for integrals. Do you recall this guy?
\( \displaystyle \int x^n \; \text{d}x = \frac{x^{n+1}}{n + 1} + C\)

Answer 13

Honestly there are so many formulas in calc2 im going nuts but vaguely yes lol

Answer 14

Oh, yea. There are a lot! Idk what learning method is easiest for you, but I heard one strategy was to use flashcards with all the formulas to get to know them well by name.

Anyways, yeah, if we apply this formula for the integral of u^(1/5), we use n=1/5.

Answer 15

so for the actual problem of 3 sq rt u du, that gives u^1/3 du, which gives -> u^ 1/3+1 divided by 3+1 + C?

Answer 16

which gives u^4/3 / 4 + C?

Answer 17

\( \large \displaystyle \color{#aaa}{\frac{d}{dt}} \int_{0}^{t^9} u^{1/3} \; \text{d} u \)
= \( \large \displaystyle \color{#aaa}{\frac{d}{dt}} \left( \frac{u^{1/3 + 1}}{1/3 + 1}+ C \right)_{0}^{t^9} \)
Be careful with those numbers! Notice the 1/3 is also in the denominator entirely.

Answer 18

crap i forgot the fraction, but yeah, so that gives u^4/3 divided by 4/3 +C, so doesnt that just cancel out?

Answer 19

The denominator and the power are not mixable. It would have to be a 4/3 multiplied on top of the fraction and on the bottom.
Instead, notice in the previous problems how there is that 5/6 u^(6/5). 5/6 is the same as 1 / (6/5). :)

Answer 20

Or in general,
\( \dfrac{a}{b} = \dfrac{1}{b/a} \)

Answer 21

so to get the 4/3 out of the denominator id multiply it out so it'd be 3/4 u ^ 4/3?

Answer 22

Yep.
Then you can plug in your a and b from F(b) - F(a). F(u) = 3/4 u ^(4/3) + C

Technically, that C is always going to be lost in the definite integration. I just show its still there but the difference will cancel out that constant.

Answer 23

so for the next step they have evaluate the f function, but they show f(t10)[the upper interval] = 5/6 t^ 12? which would translate in this problem to f(t9)= 3/4 t^ ?? where did they get 12?

Answer 24

the lower limit in the example and this problem is 0 so lost as to the 12 example

Answer 25

the exponent of u also just vanished

Answer 26

\( \displaystyle \large \color{#aaa}{\frac{d}{dt}} \left( F(t^9) - F(0) \right) \)
and we called \(\displaystyle F(u) = \frac{3}{4} u^{4/3} + C \)
So, what do you get just by plugging in u = t^9? (this is the F(t^9) part)

Answer 27

3/4(t^9)^4/3 so

Answer 28

can you simplify by exponent rules?

Answer 29

for example, \( (a^{b})^c = a^{bc} \)

Answer 30

wow i forget everything sorry

Answer 31

It 's okay. :)
It is a lot to know, that is totally understandable.

Answer 32

so 3/4 t ^12 {explains the 12 now{

Answer 33

Yep.

As for the F(0), we substitute in u = 0 into F(u).

Answer 34

just gives 0

Answer 35

3/4(0)^4/3

Answer 36

so 3/4 t^12- 0 = 3/4 t^12

Answer 37

Correct.
Then all that is left over is the derivative which was saved for last.
\( \displaystyle \frac{d}{dt} \left( \frac{3}{4} t^{1}2 - 0 \right) \)

Answer 38

uhh, \(t^{12} \)
thought I fixed it but i guess i didn't .

Answer 39

so i have to subtract out the derivative or take the derivative as the answer?

Answer 40

take the derivative for the final answer.
up until this point, the derivative was just sitting outside and all our work was inside the derivative.
but until now it had no added value by being written.

Answer 41

ok and the basic formula for that is what again

Answer 42

The power rule for derivatives is multiply by the power and then subtract one off the exponent
\( \dfrac{d}{dx} x^n = n x^{n-1} \)

Answer 43

so 3/4t^ 11? no...right?

Answer 44

You subtracted from the power correctly.
but the power also is multiplied on:
\( \dfrac{d}{dx} x^{n} = \color{red}n x^{n-1} \)

Answer 45

9t^11?

Answer 46

That looks good to me. :)

Answer 47

so just 3/4*12 gives the coefficient and then subtract off exponent right

Answer 48

Yes, that is correct.
For the derivative rule, the coefficient is multiplied first and then subtract from the power

For the integral power rule, the power is incremented by 1 first, then divided off.

Answer 49

Roger, now the second part(B) states I need to use the chain rule? Can you explain that please

Answer 50

Chain rule and also the fundamental theorem of calculus.
The formal chain rule looks a bit like this:
\( \dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx} \)
The intuitive meaning behind it is to instate a substitution to make the derivative of y with respect to u easier, while then simply multiplying on the derivative of u with respect to x.

Answer 51

Like, if you had \( y = e^{-x^2} \)
This would be hard to do as-is. But if we had \(u = -x^2\), something we can use power rule with, and then \( e^{u} \) 's derivative with respect to u is just e^u again, it becomes a multiplication of two easier derivatives. e^(-x^2) * -2x

Answer 52

As for the fundamental theorem, the statement is like this:
\( \displaystyle \dfrac{d}{dt} \int_{a}^{t} f(x) \ dx = f(t) \)

The noteworthy part is that the integral's bounds are a constant (a) and a single variable (t).

Answer 53

does this all seem familiar so far?

Answer 54

familiar but not grasped fully yes

Answer 55

Hm.. okay.
I'll try my best to make the following explanation as easy as I can.

So, our starting point is this:
\( \displaystyle \large \frac{d}{dt} \int_{0} ^{t^9} \; u^{1/3} \; du \)
I already converted the radical to u^(1/3) because it is the useful form.
By comparison to the last example, what do you notice about the boundaries that is different from the original fundamental theorem of calc I statede?

Answer 56

are you talking about the t^9?

Answer 57

Yes.
The original fundamental theorem statement is a t, not t^9.
This is where I said we should define some intermediary function, let's say w = t^9, that will take the place of the t^9.
\( \displaystyle \large \frac{d}{dt} \int_{0}^{w} u^{1/3} \; du \)
Then we are lining up to use the chain rule.

Answer 58

\(= \displaystyle \large \frac{dw}{dt} \times \frac{d}{dw} \int_{0} ^{w} u^{1/3} \; du \)

Answer 59

So now we have essentially neutralized our problem of that boundary not being t when we had d/dt.
We changed the variable to end up having w up there and d/dw. The fundamental theorem now applies, you see?

Answer 60

how did u get d/dt to dw/dt x d/dw exactly,

Answer 61

That is the chain rule step.
The easiest way I can think of understanding why is, think of the dw's as if they were fractions.

When you have the same factor in the numerator and denominator, they cancel right?

Answer 62

and in this case
\( = \displaystyle \large \frac{\cancel{dw}}{dt} \times \frac{d}{\cancel{dw}} \int_{0} ^{w} u^{1/3} \; du \)
This gets back to the last step.

Answer 63

so the chain rule was dy/dx= dy/du x du/dx... so u just took the w from the integral that replaced t and put it into?

Answer 64

Basically, yeah.
If we call that big integral thing just I, it looks like this:
\( \dfrac{dI}{dt} = \dfrac{dI}{dw} \times \dfrac{dw}{dt} \)
Multiplication is commutative, so don't mind the order being different.

Answer 65

ok slightly understand continue lol

Answer 66

Now, that whole integral thing is now in the form
\( \displaystyle \frac{d}{dt} \int_{0}^{t} f(x) dx \)
This is what all that work was done for. Because, by the fundamental theorem of calculus, we can now simplify the integral in our problem.

Answer 67

so basically that chain rule always just cancels out for us?

Answer 68

It is always set up so that it can make a cancellation, although we are going backwards from cancellation to apply chain rule.
\( \dfrac{dy}{dx} = \dfrac{du}{du} \dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \)
Apologies that this is very imprudent of me switching variables every few seconds. :p

Answer 69

not your fault. online calc2 was my own fault haha. so after this we are back to basic integral form so then?

Answer 70

So let me just put down what we had originally so we are not lost in all these variables:
\( \displaystyle \large \dfrac{dw}{dt} \times \dfrac{d}{dw} \int_{0}^{w} u^{1/3} \; du \)
The next step is applying the fundamental theorem of calculus' statement to this part:
\( \displaystyle \large \color{#aaa}{\dfrac{dw}{dt} \times } \dfrac{d}{dw} \int_{0}^{w} u^{1/3} \; du \)

Answer 71

Originally I said that the statement looked like this. This time, I will use the consistent variables I have above.
\( \displaystyle \frac{d}{d\color{green}{w}} \int_{0}^{\color{green}{w}} f(u) \; du = f(w) \)
I highlight the w's green because it is good that they are the same.

Answer 72

wait, so both ways give essentially the same answer? 9t^11

Answer 73

Yes. That's a good thing, because if we got a different answer... we had started from the same point, so we should naturally find the same end.

Answer 74

ugh i hate calc lol

Answer 75

thank you though,

Answer 76

glad to help. :p
and best of luck! it does take a lot of work to become "good" at calculus, and a lot of memorizing all those rules and becoming effective at applying them. a lot of memorizing and practice. :)