Find the derivative and simplify in each case:
$\displaystyle f(x) = arctan (x - \sqrt{1+x^2})$
$\displaystyle f(x) = e^{tan^{-1}(x^2)}$
$\displaystyle f(t) = \frac{sin^{-1}(t)}{sin^{-1}(2t)}$
Thanks to all that helps : )
i have the last 2 done and i cannot get the first one because i cannot get the derivative of the value within the brackets
and yes $\displaystyle arcsin x = sin^{-1} x$
1. ????
2. $\displaystyle f'(x) = \frac{2xe^{tan^{-1}x^2}}{1 + x^4}$
3. $\displaystyle f'(x) = \frac{\sqrt{1 - 4t^2}}{2\sqrt{1 - t^2}}$
Okay, put $\displaystyle h(x)=x-\sqrt{1+x^2}$ so that $\displaystyle h'(x)=1-\frac{x}{\sqrt{1+{{x}^{2}}}}.$
Hence, we have $\displaystyle f(x)=\arctan \big(h(x)\big)\implies f'(x)=\frac{h'(x)}{1+{{\big(h(x)\big)}^{2}}}.$ Now just remove the clutter.
-----
I didn't check if your other derivatives were okay but try this site to see it:
solving derivatives step-by-step
Hello, qzno!
The first one is challenging
. . with some surprising simplification at the end.
Of course, we use the Chain Rule . . .$\displaystyle f(x) \:=\: \arctan \left[x - \sqrt{1+x^2}\right]$
$\displaystyle f'(x) \;=\;\frac{1}{1 + \left[x-\sqrt{1+x^2}\right]^2} \cdot\bigg[1 - \frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot 2x\bigg] $
. . . . $\displaystyle = \;\frac{1}{1 + x^2 - 2x\sqrt{1+x^2} + 1 + x^2}\cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}} $
. . . . $\displaystyle = \;\frac{1}{2 + 2x^2 - 2x\sqrt{1+x^2}} \cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}} $
. . . . $\displaystyle = \;\frac{1}{2(1+x^2 - x\sqrt{1+x^2})} \cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}} $ . . . . which is what you got!
Multiply by $\displaystyle \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}$
. . $\displaystyle \frac{1}{2((1+x^2 - x\sqrt{1+x^2})} \cdot\underbrace{\frac{\sqrt{1+x^2} -x}{\sqrt{1+x^2}} \cdot\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}} $
. . $\displaystyle = \;\frac{1}{2(1+x^2 - x\sqrt{1+x^2})}\cdot\overbrace{\frac{1+x^2 - x\sqrt{1+x^2}}{1+x^2}} $ . . . now reduce!
. . $\displaystyle = \;\frac{1}{2(1+x^2)} $