Derivative of inverse trig function

[suspenc3]

Hi, can anyone check my work?

$$y=tan^-^1(x-(1+x^2)^\frac{1}{2})$$

$$y^1=\frac{1}{1+u^2}\frac{dy}{du}$$

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

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

thanks

 

[arildno]

Your post is littered with errors.
Edit: I've edited in for your forgotten square root sign.
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]<br /> Thus, we have:<br /> $$\frac{dy}{dx}=\frac{dY}{du}\mid_{u=U(x)}\frac{dU}{dx}$$<br /> Or:<br /> $$\frac{dY}{du}\mid_{u=U(x)}=\frac{1}{1+U(x)^{2}},\frac{dU}{dx}=1-\frac{x}{\sqrt{1+x^{2}}}$$<br /> Or, finally:<br /> $$\frac{dy}{dx}=(1-\frac{x}{\sqrt{1+x^{2}}})\frac{1}{1+(x-\sqrt{1+x^{2}})^{2}}$$[/itex]

 

[suspenc3]

oops..i made an error in my first equation..i edited it..can you check it now?

 

[HallsofIvy]

Hi, can anyone check my work?

$$y=tan^-^1(x-(1+x^2)^\frac{1}{2})$$

$$y^1=\frac{1}{1+u^2}\frac{dy}{du}$$

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

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

thanks

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