1. ## trig derivatives

Find the derivative and simplify in each case:

$f(x) = arctan (x - \sqrt{1+x^2})$

$f(x) = e^{tan^{-1}(x^2)}$

$f(t) = \frac{sin^{-1}(t)}{sin^{-1}(2t)}$

Thanks to all that helps : )

2. I assume that ${{\sin }^{-1}}x=\arcsin x.$

This is purely chain rule application. What have you done?

3. 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 $arcsin x = sin^{-1} x$

1. ????

2. $f'(x) = \frac{2xe^{tan^{-1}x^2}}{1 + x^4}$

3. $f'(x) = \frac{\sqrt{1 - 4t^2}}{2\sqrt{1 - t^2}}$

4. Okay, put $h(x)=x-\sqrt{1+x^2}$ so that $h'(x)=1-\frac{x}{\sqrt{1+{{x}^{2}}}}.$

Hence, we have $f(x)=\arctan \big(h(x)\big)\implies f'(x)=\frac{h'(x)}{1+{{\big(h(x)\big)}^{2}}}.$ Now just remove the clutter.

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I didn't check if your other derivatives were okay but try this site to see it:

solving derivatives step-by-step

5. haha thats what i had done! : )
thank you very much!

6. so is the answer to my first one:

$\frac{\sqrt{1+x^2} - x}{2\sqrt{1+x^2}(x^2 - x\sqrt{1+x^2} + 1)}$

or am i completely wrong? lol

7. Hello, qzno!

The first one is challenging
. . with some surprising simplification at the end.

$f(x) \:=\: \arctan \left[x - \sqrt{1+x^2}\right]$
Of course, we use the Chain Rule . . .

$f'(x) \;=\;\frac{1}{1 + \left[x-\sqrt{1+x^2}\right]^2} \cdot\bigg[1 - \frac{1}{2}(1+x^2)^{-\frac{1}{2}}\cdot 2x\bigg]$

. . . . $= \;\frac{1}{1 + x^2 - 2x\sqrt{1+x^2} + 1 + x^2}\cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}}$

. . . . $= \;\frac{1}{2 + 2x^2 - 2x\sqrt{1+x^2}} \cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}}$

. . . . $= \;\frac{1}{2(1+x^2 - x\sqrt{1+x^2})} \cdot\frac{\sqrt{1+x^2} - x}{\sqrt{1+x^2}}$ . . . . which is what you got!

Multiply by $\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}$

. . $\frac{1}{2((1+x^2 - x\sqrt{1+x^2})} \cdot\underbrace{\frac{\sqrt{1+x^2} -x}{\sqrt{1+x^2}} \cdot\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}}$

. . $= \;\frac{1}{2(1+x^2 - x\sqrt{1+x^2})}\cdot\overbrace{\frac{1+x^2 - x\sqrt{1+x^2}}{1+x^2}}$ . . . now reduce!

. . $= \;\frac{1}{2(1+x^2)}$

8. thank you!
that helped me out so much : )