Question

Differentiation....Help Required!!
(Medal+Fan)

Answer 1

Go for it.

Answer 2

Here it is @Paynesdad

Answer 3

Attachment!!

Answer 4

Ya....but not able to grasp the method!!

Answer 5

dy/dx is just 2x lol

Answer 6

Differentiate the function Y with respect to X.

Answer 7

the answer is 488/255.........how??

Answer 8

@cambrige find value of t in terms of x from the first equation and then differentiate

Answer 9

pls elaborate stepwise!!

Answer 10

Odd because I don't get 488/255 ... do you @ankit042

Answer 11

NO I have not tried...but correct method wil be finding value of t in terms of x and then differentiate

Answer 12

t(...) = x

t^2(...) = x^2

then chain rule it right?

Answer 13

but can u try it and tell if u get 488/255??

Answer 14

I know and I have done that and I did not get 488/255 ... not even close.

Answer 15

What do you get for an answer cambrige?

Answer 16

Welll first question what do you get t(x) =?

Answer 17

M stuck,like literally!!

Answer 18

Do you get an equation for t(x)?

Answer 19

nope...all I get t=2/5

Answer 20

No I can get an equation for t(x)...Are you sure the answer is 488/255?

Answer 21

yup...pretty much

Answer 22

But still u xplain ur way @Paynesdad to me!!

Answer 23

Ok here is my explanation...

Answer 24

Hang on...

Answer 25

hmm

Answer 26

\[t(1+x^2)=x\]
\[[t(1+x^2)]^2=x^2\]
\[y=[t(1+x^2)]^2+[t(x)]^2\]
\[y'=2[t(1+x^2)]~t'(1+x^2)~(2x)+2t'[t(x)]\]
\[y'=4x^2+2~t't\]

Answer 27

so stuck,eh??

Answer 28

me, yeah

Answer 29

Since \[t(1+x^2) = x \] we can deduce that\[f(x) = 1+x^2\] is the inverse of t.since t(f(x)) = x.
So let's find \[f^-1(x)\]

Answer 30

there is no guarentee that t is invertible tho or is that implied?

Answer 31

Well I guess that is an assumption that I am making.

Answer 32

(x^2 + 1) - 1 + t^2 = y

y+1-t^2 = (x^2+1)

t(y+1-t^2) = x

just thinking

Answer 33

So let me continue on my line of reasoning. If you find the inverse of f you get \[ f^-1(x)= \sqrt [x-1] \]

Answer 34

\[ f^-1(x) = \sqrt {x-1} \]

Answer 35

Now I don't know if t is invertible but I know that if you pick values of x and put them into \[t(1+x^2)\] you do get the original x back out. Now I know that is not rigorous proof but it certainly is support.

Answer 36

so this would lead me to believe that \[ t(x) = \sqrt{x-1} \}

Answer 37

\[ t(x) = \sqrt {x-1} \]

Answer 38

If we assume this is correct ...then \[t' = \frac{1}{2\sqrt{x-1}} \]

Answer 39

Now we can differentiate \[ y = x^2 + t^2\] with respect to x to get \[ \frac {dy}{dx} = 2x +2t \frac{dt}{dx} \]

Answer 40

why don't we eliminate t?

Answer 41

quite easy to get it and then take dy/dx, why not?

Answer 42

Now at x = 2 \[ t(2) = \sqrt{2-1} = 1 \] and \[ \frac{dt}{dx} = \frac{1}{2 \sqrt{2-1}}=\frac{1}{2}\]

Answer 43

So now \[ \frac{dy}{dx} = 2(2) + 2(1) \left( \frac{1}{2} \right) = 4+1 =5\]

Answer 44

Elinimate t starting from where ? Starting from the assumption that \[t=\sqrt{x-1}\]? @cambrige

Answer 45

yup

Answer 46

connectivity lost so

Answer 47

If so you get the same value ... 5!

Answer 48

This is why I am questioning whether the answer that you have been provided is correct.

Answer 49

Eliminate from \[t(1+x^2)=x \rightarrow t = \frac{x}{1+x^2}\rightarrow t^2 = \frac{x^2}{(1+x^2)^2}\]
and for y , t^2 = y-x^2
then just make them equal, right?

Answer 50

I was assuming that t was a function so t(1+x^2) is a function notation nott time (1+x^2) ... but let me think about what you are saying for a minute.

Answer 51

surely x,y are functions w.r.t t. So parameter equation by eliminating t is reasonable.

Answer 52

@Loser I am saying I was assuming that t(1+x^2) was a function notation not a multiplication statement... let me work through this.

Answer 53

@cambrige you still there?

Answer 54

@Loser66 I think you are on to something. If you are correct and the equation \[t(1+x^2) = x\] means \[t * (1+x^2) = x\] and not that t of (1+x^2) = x (the function notation) then I get that dy/dx = 488/125 which is very similar to the answer that @cambrige said it was supposed to be. Although he said 488/255 which was perhaps a typo.

Answer 55

Ya...so must be a typo!!.....and multiplication....whew finally!!

Answer 56

can u give the steps @Paynesdad ??

Answer 57

get t,diff it,sub x=2??

Answer 58

So @cambrige if you come back to this problem I see two possible answers depending on what is meanth by \[t(1+x^2)=x\] . If that is meant to say that the function t at 1 + x^2 is equal to x then that would imply that the function t is such that \[t(x)= \sqrt{x-1}\] and thus \[ \frac{dt}{dx}= \frac{1}{2\sqrt{x-1}}\].

So for when we differentiate, \[y= x^2 + t^2\], we get \[\frac{dy}{dx} =2x + 2t \frac{dt}{dx}\].

When x =2, \[t=\sqrt{2-1}=1\] and \[\frac{dt}{dx}=\frac{1}{2\sqrt{2-1}}=\frac{1}{2}\].
so \[\frac{dy}{dx}=2(2) + 2(1) \left( \frac{1}{2} \right) = 4+1 = 5\]

Answer 59

thanks for the valuable insight @Loser66

Answer 60

don't understand what you mean. I did nothing

Answer 61

@cambrige but if we use what @Loser66 suggested we get much closer to what you stated the answer to be.

Answer 62

U helped us realize there was a multilication possibility!!

Answer 63

I get 148/125 which is similar and perhaps there was a typo in the answer. Here is my work for that...

Answer 64

my simple brain cannot understand what you are doing. hehehe. sorry

Answer 65

thanks to @ankit042 ,@Futuremath!!

Answer 66

Again if we differentiate \[y=x^2+t^2\], we still get \[\frac{dy}{dx}= 2x+2t \frac {dt}{dx}\]

Answer 67

I am thankin u Loser,nthin else to think!!

Answer 68

Last but not the least THANK YOU
@Paynesdad for ur help!!

Answer 69

Are u posting ur work @Paynesdad ??

Answer 70

But now if \[t* (1+x^2) = x\] then \[t=\frac{x}{1+x^2} \] then \[\frac{dt}{dx} = \frac{1-x^2}{(1+x^2)^2}\]
at x = 2
\[t = \frac{2}{1+2^2} = \frac {2}{5}\]

Answer 71

I posted the one thought above and Loser66's idea now

Answer 72

Continuing
if x = 2 then \[ \frac{dt}{dx} = \frac{1-2^2}{(1+2^2)^2} = -\frac{3}{25}\]
and so \[ \frac {dy}{dx}= 2x+2t \frac{dt}{dx} = 2(2) + 2 \left( \frac{2}{5} \right) \left( -\frac{3}{25} \right) = 4 - \frac{12}{125} = \frac{500}{125} - \frac{12}{125} = \frac{488}{125} \]

Answer 73

@cambrige does that make sense? I can not see how 488/255 could be an naswer but I am confident either 5 or 488/125 is correct based on what was intended in the problem statement. It was rather ambiguous.

Answer 74

Ya so I got stuck at 2/5!!

Answer 75

Well you were on the right track!

Answer 76

kinda but stuck!!

Answer 77

so dats it,I guess.....tnx again!!