# Thread: How to differientiate these Polynomial functions

1. Originally Posted by ghostanime2001
HOly **** !?!?!?!?!?!??!?! You almost finished my quest though T_T

But just one more question is it possible to take out the negative sign out before getting to that step so the approach to the answer is more clear and logical or is that it ??
if by "take out the negative sign" you mean "factor out -1" then, uh, yeah, you can do that whenever. whatever looks best to you.

Btw... how do u guys get that text i want to like make my own online source of solving homework questions and post it on the web so its a whole lesson from what i've learned posted on the net for the rest of the world. I'm just wondering cuz that text you guys wrote, all those mathematical symbol is so clear and nice i wanna know if you can do that in Word or something or possibly download a program that types that text. Thanks
to create documents to use elsewhere, you can download programs like MikTex and TexNicCenter. for use in the forum, we have LaTeX built in, so you type things between  tags and it shows up like that. see our LaTeX forum to learn how to do it. you will need to know LaTeX coding to use the afore mentioned programs as well

2. Originally Posted by ghostanime2001
If i differentiate using your first way Mathstud28
y = (1- √x)²
= (1- √x)(1- √x)
= 1 - 2√x + x

f'(x) = -2(1/2)x^-½ + 1
= -x^-½ + 1
= -1/√x + 1

substituting x->a and y'->? i get this

? = -1/√a + 1

How do i from here show the slope of tha tangent at that √x + √y = 1 at P(a,b) is

Well then do it the second way I showed you

$\color{red}\frac{1}{2\sqrt{x}}+\frac{y'}{2\sqrt{y} }\Rightarrow{y'=\frac{-\sqrt{y}}{\sqrt{x}}}$

so

$\color{red}y'(a,b)=\frac{-\sqrt{b}}{\sqrt{a}}$

I think we can finally put this to rest.
-√b
-----
√a

The thing to remember what im sayin here is that this worksheet i got in school, was given before the class learned implicit differentiation. So That's my problem here how do i prove it equals that without having to even think about implicit differentiation, if i was a currently enrolled calc student in class that didnt know what implicit differentiation meant.

Then what you will have to do is find a way to relate what you got by normal differentiation back to your original function

Thanks again
EDIT: Sorry, completely missed the existence of a second page

3. How do i solve the the second question then ? my first post to this threat.

4. Originally Posted by ghostanime2001
How do i solve the the second question then ? my first post to this threat.
you know how to find a tangent line to a curve at a certain point, right? just find the line. then you will be able to see what the x-intercept is, and how it changes as n becomes bigger

5. Originally Posted by Jhevon
you know how to find a tangent line to a curve at a certain point, right? just find the line. then you will be able to see what the x-intercept is, and how it changes as n becomes bigger
I'm tempted to ask what part of my reply regarding Q2 (all those miles back at post #2) the OP does not understand ......

6. This is the second question btw i just want to post what i got here so i know im not making a mistake anywhere.

f(x) = x^n
f '(x) = nx^n-1 P(1,1)

y-y1 = m(x - x1)
y - 1 = nx^n-1 (x - x1)

and now im stuck. LOL. I just know that im supposed to put the equation in standard form and then make y = 0 and solve for x since that would give me the x-intercept. As for the second part about n-->+OO i dont i think im supposed to take the limit. So i can't go any furthur just yet without working my way through from the top part.

7. Originally Posted by mr fantastic
I'm tempted to ask what part of my reply regarding Q2 (all those miles back at post #2) the OP does not understand ......
Oh, i'm sorry Mr F. i didn't see that you addressed the problem. my bad ....i feel like such a New Yorker saying that

8. Originally Posted by ghostanime2001
This is the second question btw i just want to post what i got here so i know im not making a mistake anywhere.

f(x) = x^n
f '(x) = nx^n-1 P(1,1)

y-y1 = m(x - x1)
y - 1 = nx^n-1 (x - x1)

and now im stuck. LOL. I just know that im supposed to put the equation in standard form and then make y = 0 and solve for x since that would give me the x-intercept. As for the second part about n-->+OO i dont i think im supposed to take the limit. So i can't go any furthur just yet without working my way through from the top part.
Have you actually read what I said in post #2 re: Q2?

9. yes but i dont understand why m=n ?? and what happened to the ^n-1 in the point slope equation ? why is it left out ?

10. Originally Posted by ghostanime2001
yes but i dont understand why m=n ?? and what happened to the ^n-1 in the point slope equation ? why is it left out ?
By convention, we usually use $m$ to represent the slope of a line (in this case, the tangent line). And as Mr. F said, since $f'(x) = nx^{n-1},$ we will have $f'(1) = n,$ so the tangent line has a slope of $n$ when $x = 1$. Then you can write an equation for the tangent line using point-slope form, like so:

$(y - y_0) = m(x - x_0)\Rightarrow y - 1 = n(x - 1)$

11. Originally Posted by ghostanime2001
yes but i dont understand why m=n ?? and what happened to the ^n-1 in the point slope equation ? why is it left out ?
Just to save going aaaaall the way back there, here's the question:

Originally Posted by ghostamime2001
[snip]
2. For the power functions f(x) = x^n, find the x-intercept of the tangent to its graph at point (1,1). What happens to the x-intercept as n increases without bound (n --> +OO (positive infinity) ) ? Explain the result geometrically.
[snip]
And here's what I said:

Originally Posted by mr fantastic
[snip]
$f'(x) = n x^{n-1} \Rightarrow m = n \, \text{at} \, x = 1$. Then the equation of the tangent is $y - 1 = n(x - 1) \Rightarrow y = ......$
[snip]
Now I want to establish what you do and don't understand:

Do you understand that:

* f'(x) gives the gradient of the tangent to a curve?

* m is commonly used as a symbol for gradient?

* you substitute x = 1 into $f'(x) = n x^{n-1}$ to get the gradient of the tangent to the curve at x = 1?

* $1^{n-1} = 1$?

12. ummm almost done but how do i differentiate piecewise functions ? I've done the graphing on paper and i dont know how to graph it on the site to show you but the least i can say is i've graphed the piecewise function but now i need to differentiate it. I've differentiated the little pieces in the piecewise function but the intervals somehow change when i differentiate it. I don't understand that. What i mean is when something in a piecewise function is

x+6 x >=3 which is normal part inside piecewise function then when i differentiate it the interval becomes:

x+6 x > 3 what happened to the x being greater or equal to 3 after i differentiated it ? how does that vanish after differentiating that little part in the entire piecewise function and if u dont know what function im talking about look at my first post on this threat and that x+6 x>3 i got from my answer sheet but i dont know why or how it changes

13. Originally Posted by ghostanime2001
ummm almost done but how do i differentiate piecewise functions ? I've done the graphing on paper and i dont know how to graph it on the site to show you but the least i can say is i've graphed the piecewise function but now i need to differentiate it. I've differentiated the little pieces in the piecewise function but the intervals somehow change when i differentiate it. I don't understand that. What i mean is when something in a piecewise function is

x+6 x >=3 which is normal part inside piecewise function then when i differentiate it the interval becomes:

x+6 x > 3 what happened to the x being greater or equal to 3 after i differentiated it ? how does that vanish after differentiating that little part in the entire piecewise function and if u dont know what function im talking about look at my first post on this threat and that x+6 x>3 i got from my answer sheet but i dont know why or how it changes
The intervals do not change other than for points on the boundaries between the peices where the derivative may not exist because the derivative from one side is not equal to the derivative from the other, or there may be a jump discontinuity at the boundary.

You have f(x) = x+6 when x>=3, and lets say f(x) = 9-x^2 when x<3

Then f'(x)=1 when x>3 (that's the derivative of x+6), and f'(x)=-2x when x<3 (that's the derivative of 9-x^2). The derivative at x=3 does not exist because 1 != -2(3) that is the derivative from the left does not equal the derivative from the right (note there is no jump discontinuity here, but if there had been the derivative would be undefined at the join even if the derivatives from the left and right had been equal).

RonL

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