Question

How do I do this type of questions?
Calc 1

Answer 1

It says, find f '(x), if\[\large f(x)=\int\limits_{4x}^{1}~\cot^2(t)~dx\]
Can someone guide me through the problem please?

Answer 2

@wio , can you help please? sorry for bothering you....

Answer 3

I know how to integrate cot^2t, but what do I have to do, can you guide me please?

Answer 4

Well, suppose that: \[
g'(t) = \cos^2(t)
\]

Answer 5

okay.

Answer 6

That would mean that: \[
f(x) = \int_{4x}^1g'(t)~dt=g(1)-g(4x)
\]

Answer 7

yes.

Answer 8

Therefore: \[
f'(x) = \left(g(1)-g(4x)\right)'
\]

Answer 9

Can you do it now?

Answer 10

Let's see, so first I integrate.... wait, I don't have to integrate, I can tell it is
-cot^2(4x), right?

Answer 11

No.

Answer 12

because the g(1) is constant, so the derivative will make it a zero.

Answer 13

We don't need to find \(g(t)\).

Answer 14

I have to integrate, yes?

Answer 15

No, you don't need to integrate. You need to differentiate.

Answer 16

find the derivative of cot^2(t) ?

Answer 17

Do you need help differentiating here? \[
f'(x) = \left(g(1)-g(4x)\right)'
\]

Answer 18

I don't understand your set up so well.

Answer 19

Does this make sense? \[
f'(x) = \frac{d}{dx}f(x) = \frac{d}{dx}\left(g(1)-g(4x)\right)
\]

Answer 20

\[\frac{ d }{ dx }\left(\begin{matrix} \cot^2(1)-\cot^2(4x)\end{matrix}\right)\]Like this?

Answer 21

Okay, we don't know what \(g(t)\) is. We can simplify without knowing what it is.

Answer 22

First step:\[
\frac{d}{dx}\left(g(1)-g(4x)\right) = \frac{d}{dx} g(1) - \frac{d}{dx}g(4x)
\]

Answer 23

Next step:\[
= - \frac{d}{dx}g(4x)
\]

Answer 24

So my set up,
\[\frac{ d }{ dx }\left(\begin{matrix} \cot^2(1)-\cot^2(4x)\end{matrix}\right)\]
was correct?

Answer 25

It was wrong... that's what I've been saying.

Answer 26

Sorry

Answer 27

I am stupid.

Answer 28

Do you know how you would simplify this:\[
- \frac{d}{dx}g(4x)
\]

Answer 29

IF I knew the g(x), what is it?

Answer 30

You don't really need to know. Let's just say \(u=4x\) to make things simpler.
Could you do this? \[
\frac{d}{dx}g(u)
\]

Answer 31

g'(u) ?

Answer 32

No, you need to remember the chain rule.

Answer 33

\[\frac{ dy }{ dx }=\frac{ dy }{ du }~\frac{ du }{ dx }\]this?

Answer 34

Sure.

Answer 35

In our case: \[
\frac{d}{dx}g(u)=\frac{dg}{du} \frac{du}{dx} =g'(u) (4x)' = g'(4x)(4x)'
\]

Answer 36

Now, we said that: \[
g'(t) = \cot^2(t)
\]

Answer 37

That means that: \[
g'(4x)=\ldots
\]

Answer 38

cot^2(4x).... whoo

Answer 39

the chain rule for the 4x....

Answer 40

yes..

Answer 41

\[
f'(x) =- 4\cot^2(4x)
\]

Answer 42

The \((4x)' = 4\).

Answer 43

Yes, it is -, because it was g(1)-g(4x)
Not like I am correcting you or anything like that. I think I kind of understand it, but still really poorly.

Answer 44

I'll read some tutorials online and my notes, I guess I can better understand it then. Tnx for the time and effort.

Answer 45

The only problem here is that you're not completely comfortable with your calculus as an algebraic operation.

Answer 46

The implicit differentiation part.

Answer 47

Perhaps, that is it.
Like I know just how to "calculate" but for the rest my head is not working-:(

Answer 48

It is just harder for me to understand the query of the querstion....
well, tnx again:)

Answer 49

I'll do the problem once more quickly so you can review.

Answer 50

that's very nice;)

Answer 51

Here is the entire thing. You can look at each step and let me know if there is a step that you just don't get.

Answer 52

Okay... will see.

Answer 53

\[
\begin{split}
f'(x)
&= \frac{d}{dx}f(x)&\quad&\\
&= \frac{d}{dx}\int_{4x}^1\cot^2(t)~dt&&\text{let }g'(t) = \cot^2(t)\\
&= \frac{d}{dx}\int_{4x}^1g'(t)~dt\\
&= \frac{d}{dx}(g(1)- g(4x))\\
&= \frac{d}{dx}g(1)- \frac{d}{dx}g(4x)\\
&= - \frac{d}{dx}g(4x)&&\text{let }u=4x\\
&= - \frac{d}{dx}g(u)\\
&= - g'(u)u'(x)&&\text{let } \cot^2(t)=g'(t)\\
&= - \cot^2(u)\frac{d}{dx}u&&\text{let } 4x=u\\
&= - \cot^2(4x)\frac{d}{dx}4x\\
&= - 4\cot^2(4x)

\end{split}

\]

Answer 54

ohh, so the "let y = 4x" shouldn't be there?

Answer 55

ohh, you fixed to u=4x... I se

Answer 56

The y is next to u key.

Answer 57

Yes yes:)

Answer 58

u'(x) is the chain rule. Okay I think I am getting it.

Answer 59

and you mean let g'(u) = cos^2(u).... right?

Answer 60

I mean 4th to the last line...

Answer 61

Yes, it is sort of the same thing.

Answer 62

Yes, ty, I get it. I don't promise that I'll be able to do another one like this on my own, but I appreciate your help!

Answer 63

I'll take a note of it on my papaer.

Answer 64

There is a general formula for these things.

Answer 65

So this is all I should have done?

Answer 66

\[
\frac{d}{dx}\int_{a(x)}^{b(x)} f'(t)~dt = f'(b(x))b'(x)-f'(a(x))a'(x)
\]

Answer 67

yeah... I msee what you changed.

Answer 68

the derivative, chain rule for each, right? for f(b) and f(a).

Answer 69

Well, the thing is, it isn't that complicated to begin with, it's just when I do it with such detail, then it becomes very long.

Answer 70

Yeah... I mean I am just very bad at problems that aren't like
"differentiate f(x)" you know what I mean( ?)

I won't necessarily know the "steps" but I am a pretty good calculator.

Answer 71

You could have said: \[
\frac{d}{dx}\int_{4x}^1\cot^2(t)~dt = \cot^2(1)(1)' - \cot^2(4x)(4x)' = -4\cot^2(4x)
\]But you might not see where that comes from.

Answer 72

I see why is that (sort of).

Answer 73

yes, I think I got it....tnx