Question

How to solve!
Help pleaseee!
Find dy/dx of the problems below.

Answer 1

\[y= \int\limits_{\tan x}^{0} \frac{ dt }{ 1+ t^2 }\]
\[y=\int\limits_{0}^{x^2} (\cos \sqrt{t}) dt\]

Answer 2

`Fundamental Theorem of Calculus, Part 1`:\[\large \frac{d}{dx}\int\limits_c^x f(t)\;dt \qquad=\qquad f(x)\]

Remember this thing? :)
We'll have to use this, it'll just be a tad bit more complex.

Answer 3

yes, but I don't know how to apply it to the problem very well..

Answer 4

For the first problem, I'm going to write it in a simplified form for a sec.
Maybe that will help you to understand what's going on.

\[\large y=\int\limits\limits_{\tan x}^{0} f(t)\;dt\]Integrating gives us,\[\large y=F(t)|_{\tan x}^0 \qquad =\qquad F(0)-F(\tan x)\]
Taking the derivative gives us,\[\large y'=0-f(\tan x)\color{royalblue}{(\tan x)'}\]
Where f represents our original function that was in the integral,
except instead of t's, we have (tan x)'s replacing all of our t's now.
Understand the blue term?
That shows up due to the chain rule.

Try to digest this a sec :) Lemme know what part is confusing.
*F(t) represents the antiderivative of f(t).

Answer 5

so the blue represents the derivative of t? as according to the chain rule?

Answer 6

@zepdrix

Answer 7

The blue represents the derivative of the thing we plugged in for t, yes.

After integrating we got some ugly function which we call \(\large F(t)\).
We then plug in our limits.
We then take the derivative of our ugly function.

\(\large \frac{d}{dx}F(\tan x)=f(\tan x)\frac{d}{dx}\tan x\)

Taking the derivative of our ugly function gives us back \(\large f(t)\), but now it's f(stuff), since we've plugged in our limits.

We have to apply the chain rule, multiplying by the derivative of the (stuff).

Answer 8

so we get
\[\frac{ d }{ dx } F(\tan x)= f(\tan x)(\sec^2 x)\]
@zepdrix

Answer 9

Integrating gave us this,
\[\large y=F(0)-F(\tan x)\]See how the thing we `care about` is being evaluated at the lower limit?
Meaning: It has a subtraction sign on it.

Yes what you've done so far looks good! :)
Let's not forget about that subtraction sign now in the next step.
Taking the derivative gives us,\[\large y'=0-f(\tan x)(\sec^2x)\]

Answer 10

Just so we're clear on what f(t) means, here is how it will change the function:
\[\large f(\color{green}{t})=\frac{1}{1+\color{green}{t}^2} \qquad\to\qquad f(\color{green}{\tan x})=\frac{1}{1+\color{green}{(\tan x)}^2}\]

Answer 11

So we determined that our derivative will give us this:
\[\large y'=-(\sec^2x)\cdot f(\color{green}{\tan x})\]

Answer 12

ohhhh, okay. and that's dy/dx right?

Answer 13

Yes, dy/dx is y'.
I'm lazy with notation sometimes lol.

Answer 14

haha. I caught on!(: what about the second function?

Answer 15

Mmm let's use this same approach. Hopefully the function notation is making some sense to you.

\[\large y=\int\limits\limits_{0}^{x^2} f(t)\;dt \qquad=\qquad F(x^2)-F(0)\]

Taking the derivative, applying the chain rule where needed, gives us,\[\large y'=f(x^2)\color{royalblue}{(x^2)'}-0\]

Answer 16

\[\large y'=2x \cdot f(x^2)\]

Answer 17

that one was a lot easier then I thought. haha. thank you!!!

Answer 18

Just make sure that for your answers, you're actually writing out the functions. Don't simply leave it as f(x^2), hehe.

\[\large y'=2x\cdot \cos\sqrt{x^2}\]

Answer 19

so for the first one how do i evaluate f(tanx)? or does it just remain that?

Answer 20

I posted it above silly +_+\[\large f(\color{green}{t})=\frac{1}{1+\color{green}{t}^2} \qquad\to\qquad f(\color{green}{\tan x})=\frac{1}{1+\color{green}{(\tan x)}^2}\]
You would want to include that in your answer ~ the part on the far right. Fill that in for your f(tan x).

Answer 21

haha I caught tht right whn i asked the question haha..

Answer 22

XD

Answer 23

it would be
\[y'=-\frac{ \sec^2x }{ 1+(\tan x)^2 }\]
just to double check lol

Answer 24

Mmmm yah that looks good!

Answer 25

Your teacher might want you to simplify it down further, I'm not sure though.

Do you see how we could simplify that one?

Answer 26

Simplify?

Answer 27

This is one of our `Square Identities` back from Trig:
\[\large 1+\tan^2x=\sec^2x\]

Answer 28

so it would just be -1?

Answer 29

Ya :)

Answer 30

ohhhhh! haha(: Thank you!
could you by chance help me with another integral problem? I seriously suck at them ;/ @zepdrix

Answer 31

And for the other problem, you could write it like this I suppose,
\[\large y'=2x\cdot \cos\sqrt{x^2} \qquad\to\qquad y'=2x\cdot \cos|x|\]That one isn't too important though.

Sure :U Open it in a new thread.