- #1

- 284

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I tryed a lot and got this...

Can you spot any mistakes or give me hints on how to aproach this:yuck:

Thanks

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- Thread starter Focus
- Start date

- #1

- 284

- 3

I tryed a lot and got this...

Can you spot any mistakes or give me hints on how to aproach this:yuck:

Thanks

- #2

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- 0

- #3

- 284

- 3

I am so confused...I know that the second one can be obtained by using y=arsinh(x) but I cant get the log of the function to differentiate to that:uhh:SebastianG said:

- #4

HallsofIvy

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[tex]y= ln\left[x+ (1+x^2)^{\frac{1}{2}}\right][/tex]

then

[tex](1+x^2)\left(\frac{dy}{dx}\right)^2= 1[/tex]

which is the same as saying that

[tex]\frac{dy}{dx}= \frac{1}{(1+x^2)^{\frac{1}{2}}}[/tex]

- #5

- 284

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The problem is I can't derive [tex]\frac{dy}{dx}= \frac{1}{(1+x^2)^{\frac{1}{2}}}[/tex] from [tex]y= ln\left[x+ (1+x^2)^{\frac{1}{2}}\right][/tex]...HallsofIvy said:

[tex]y= ln\left[x+ (1+x^2)^{\frac{1}{2}}\right][/tex]

then

[tex](1+x^2)\left(\frac{dy}{dx}\right)^2= 1[/tex]

which is the same as saying that

[tex]\frac{dy}{dx}= \frac{1}{(1+x^2)^{\frac{1}{2}}}[/tex]

I can only derive it from y=arsinh(x)

- #6

NateTG

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Focus said:The problem is I can't derive [tex]\frac{dy}{dx}= \frac{1}{(1+x^2)^{\frac{1}{2}}}[/tex] from [tex]y= ln\left[x+ (1+x^2)^{\frac{1}{2}}\right][/tex]...

I can only derive it from y=arsinh(x)

It seems like it pops right out from the chain rule.

- #7

HallsofIvy

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[tex]y= ln\left[x+ (1+x^2)^{\frac{1}{2}}\right][/tex]so

[tex]y'= \frac{1}{x+(1+x^2)^{\frac{1}{2}}}\left(1+(\frac{1}{2})(1+ x^2)^{-\frac{1}{2}}(2x)\right)\)[/tex]

Now factor that

[tex](1+ x^2)^{-\frac{1}{2}}[/tex]

out of the numerator and every thing else cancels!

- #8

- 284

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Hmmm...I tried that but I'll do it again, cheers for the help

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