Derivative of inverse trig function

suspenc3
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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
 
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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 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}}
 
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oops..i made an error in my first equation..i edited it..can you check it now?
 
suspenc3 said:
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).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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