# 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

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Homework Helper
Gold Member
Dearly Missed
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]
Thus, we have:
$$\frac{dy}{dx}=\frac{dY}{du}\mid_{u=U(x)}\frac{dU}{dx}$$
Or:
$$\frac{dY}{du}\mid_{u=U(x)}=\frac{1}{1+U(x)^{2}},\frac{dU}{dx}=1-\frac{x}{\sqrt{1+x^{2}}}$$
Or, finally:
$$\frac{dy}{dx}=(1-\frac{x}{\sqrt{1+x^{2}}})\frac{1}{1+(x-\sqrt{1+x^{2}})^{2}}$$

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suspenc3
oops..i made an error in my first equation..i edited it..can you check it now?

Homework Helper
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).