# Derivative of inverse trigonometric functions: $f(x)=\arcsin\left(2x\sqrt{1-x^2}\right)$

While trying to find the derivative of $$f(x)=\arcsin\left(2x\sqrt{1-x^2}\right)$$ we arrive at two different answers by substituting $x=\sin t$ or $x=\cos t$. The aim is to simplify it as $\arcsin(\sin2t)$ and further simplify it as $2t$ and take its derivative. Proceeding in that manner, I arrived at two different answers for the derivative. I could understand that it was because the function $f$ is defined differently in different intervals, and as I have solved the problem without considering the intervals, I hit this answer. Also, I thought that since $\sin t$ and $\cos t$ will coincide only when $t=\dfrac{\pi}4$, $x=\pm\dfrac1{\sqrt{2}}$ will be the points where the definition changes. I got some more clarity when I looked at the graph of $f(x)$ but I could not figure out a way to systematically figure out what answer suits what interval.

MAJOR EDIT: I made a mistake in the function itself that I had given. My question earlier read: $$f(x)=\arcsin\left(2x\sqrt{x^2-1}\right)$$ But I wanted only for what I've now altered it as. My sincere apologies...

EDIT: My question is how to figure out the interval in which the substitution applies for such problems in general. I don't need the expression for the derivative of $f(x)$.

Note: I am new to this community, so please point out any deviation from the policy, if I have deviated. Also, I couldn't post my working as I do not have $10$ reputation.

• At least put each formula between dollars. – Nosrati Aug 11 '18 at 4:01
• The MathJax tutorial may help. – Xander Henderson Aug 11 '18 at 4:02
• Thank you for the link... I've made the changes. – Raghavasimhan Varadharajan Aug 11 '18 at 4:10
• your substitution with $sin$ or $\cos$ doesn't work. $\sqrt{\sin^2x-1}=\sqrt{-\cos^2x}$ is undefined . . . – Nosrati Aug 11 '18 at 4:21
• $x^2-1\geq0$ shows $|x|\geq1$ while $|\sin|\leq1$ – Nosrati Aug 11 '18 at 4:34

$$f(x)=\arcsin(2x\sqrt{x^2-1})$$ $$\sin(f(x))=2x\sqrt{x^2-1}$$ $$f'(x)\ \cos(f(x))=2\sqrt{x^2-1}+2x\dfrac{2x}{2\sqrt{x^2-1}}$$ $$f'(x)\sqrt{1-\sin^2f(x)}=2\dfrac{2x^2-1}{\sqrt{x^2-1}}$$ $$f'(x)=2\dfrac{2x^2-1}{\sqrt{x^2-1}\sqrt{1-4x^4+4x^2}}$$

• Thank you for the answer! But my question was about the answer I got in a certain method. It would be of great help if you could clarify that question – Raghavasimhan Varadharajan Aug 11 '18 at 4:16
• Really I couldn't read your post. Now you've edited so I read it! – Nosrati Aug 11 '18 at 4:18
• oh yes. And thanks for your suggestion in the comments. – Raghavasimhan Varadharajan Aug 11 '18 at 4:19

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

Replace x=cosh$\theta$

then f(x) = arcsin(sinh$2\theta$)

by differentiating f(x),

f '(x) = $\frac{2cosh(2\theta)}{\sqrt{1-sinh^2(2\theta)}\sqrt{x^2-1}}$

Replacing $cosh(2\theta) = 2x^2-1$

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

• Is x=sinhθ a valid substitution? – Raghavasimhan Varadharajan Aug 11 '18 at 4:24
• @RaghavasimhanVaradharajan, x = $cosh(\theta)$ is a valid substitution – Narendra Deconda Aug 11 '18 at 5:08