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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Inside Out on October 12, 2010, 11:44:22 pm

Title: Another hard question
Post by: Inside Out on October 12, 2010, 11:44:22 pm: If y=(x+(x²+1)^1/2)², show dy/dx=2y/(x²+1)^1/2
Title: Re: Another hard question
Post by: brightsky on October 12, 2010, 11:47:24 pm: dy/dx? ;)

Again directly apply the chain rule:

$\frac{dy}{dx} = \frac{1}{2} (x^2 + x + 1)^{-\frac{1}{2}} (2x + 1)$

Then substitute $y = (x^2 + x + 1)^{\frac{1}{2}}$ into there and simplify.

EDIT:...wait wat?
Title: Re: Another hard question
Post by: TrueTears on October 12, 2010, 11:48:54 pm: question is wrong as $1 \neq RHS$

assuming dy/dx

dy/dx = (2x+1)/(2y)

dx/dy = 2y/(2x+1)
Title: Re: Another hard question
Post by: Inside Out on October 12, 2010, 11:53:24 pm: QUesiton has been modified :D
Title: Re: Another hard question
Post by: brightsky on October 13, 2010, 12:16:55 am: In that case, emulate the same process:

$\frac{dy}{dx} = 2\left(x + (x^2 + 1)^{\frac{1}{2}}\right) \frac{d}{dx}\left(x + (x^2 + 1)^{\frac{1}{2}}\right)$

Finding $\frac{d}{dx}\left(x + (x^2 + 1)^{\frac{1}{2}}\right) = 1 + \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}2x$

Substituting in:

$2\left(x + (x^2 + 1)^{\frac{1}{2}}\right) \left(1 + \frac{1}{2}(x^2 + 1)^{-\frac{1}{2}}2x\right)$

$= 2\left(x + (x^2 + 1)^{\frac{1}{2}}\right) \left(1 + \frac{x}{(x^2 + 1)^{\frac{1}{2}}}\right)$

$= 2 \left(x + \frac{x^2}{\sqrt{x^2 + 1}} + \sqrt{x^2 +1} + x\right)$

$= 2 \frac{x\sqrt{x^2 + 1} + x^2 + x^2 + 1 + x\sqrt{x^2 + 1}}{\sqrt{x^2 + 1}}$

$= 2 \frac{2x\sqrt{x^2 + 1} + 2x^2 + 1}{\sqrt{x^2 + 1}}$

Since $y = (x + \sqrt{x^2 + 1})^2 = x^2 + 2x\sqrt{x^2 + 1} + x^2 + 1 = 2x^2 + 2x\sqrt{x^2+1} + 1$

Then the expression becomes:

$\implies 2 \frac{y}{\sqrt{x^2 + 1}}$

$= \frac{2y}{\sqrt{x^2 +1}}$
Title: Re: Another hard question
Post by: 98.40_for_sure on October 13, 2010, 12:22:07 am: $y=(x+\sqrt{x^{2}+1})^{2}$

let $u=x+\sqrt{x^{2}+1}$

$\frac{du}{dx}=1+\frac{x}{\sqrt{x^{2}+1}}$

$y=u^{2};\frac{dy}{du}=2u$

$\frac{dy}{dx}=2u(1+\frac{x}{\sqrt{x^{2}+1}})$

$=2u+\frac{2ux}{\sqrt{x^{2}+1}}$

$=\frac{2(x+\sqrt{x^{2}+1)}(\sqrt{x^{2}+1})+2x(x+\sqrt{x^{2}+1})}{\sqrt{x^{2}+1}}$

$=\frac{2x\sqrt{x^{2}+1}+2(x^{2}+1)+2x^{2}+2x\sqrt{x^{2}+1}}{\sqrt{x^{2}+1}}$

$=\frac{2(2x\sqrt{x^{2}+1}+(x^{2}+1)+x^{2})}{\sqrt{x^{2}+1}}$

$=\frac{2(x+\sqrt{x^{2}+1})^{2}}{\sqrt{x^2}+1}$

$=\frac{2y}{\sqrt{x^{2}+1}}$ as required

EDIT: DAMMIT BEATEN :(
Title: Re: Another hard question
Post by: Inside Out on October 13, 2010, 10:25:37 pm: ahhhhhhhhhhh i get it thanks vn people! :D