Derivative of inverse trig function

  • Thread starter suspenc3
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  • #1
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Hi, can anyone check my work?

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

[tex]y^1=\frac{1}{1+u^2}\frac{dy}{du}[/tex]

let u = (x-(1+x^2))

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

thanks
 
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Answers and Replies

  • #2
arildno
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Your post is littered with errors. :mad:
Edit: I've edited in for your forgotten square root sign. :mad::mad:
Here's how you may do it:
Set [itex]Y(u)=\tan^{-1}(u),U(x)=(x-\sqrt{1+x^{2}}), y(x)=Y(U(x))[/tex]
Thus, we have:
[tex]\frac{dy}{dx}=\frac{dY}{du}\mid_{u=U(x)}\frac{dU}{dx}[/tex]
Or:
[tex]\frac{dY}{du}\mid_{u=U(x)}=\frac{1}{1+U(x)^{2}},\frac{dU}{dx}=1-\frac{x}{\sqrt{1+x^{2}}}[/tex]
Or, finally:
[tex]\frac{dy}{dx}=(1-\frac{x}{\sqrt{1+x^{2}}})\frac{1}{1+(x-\sqrt{1+x^{2}})^{2}}[/tex]
 
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  • #3
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oops..i made an error in my first equation..i edited it..can you check it now?
 
  • #4
HallsofIvy
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suspenc3 said:
Hi, can anyone check my work?

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

[tex]y^1=\frac{1}{1+u^2}\frac{dy}{du}[/tex]

let u = (x-(1+x^2))

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

thanks
If you edited this after arildno looked at it, you are still missing a square root:
[tex]y=tan^-^1(x-(1+x^2)^\frac{1}{2})= tan^{-1}(u)[/tex]
for u= (x- (1+ x2)1/2).
 

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