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taking JDs advice i used $(fg)'=f'g+fg'$ rule $$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$ $$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$ so anyway adding together we get $$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt{1-x^2}}$$ which can be simplified $$\frac {sin^{-1}x}{(1-x)^3}+\frac x{\sqrt{(1-x^2)^2}}$$ i then multiplied $$\frac x{\sqrt{(1-x^2)^2}}$$ with $\sqrt{(1-x^2)}$ on both numerator and denominator getting $$\frac {x(\sqrt{(1-x^2)})}{\sqrt{(1-x^2)^3}}$$ and combining them both we get $$\frac{sin^{-1}x+x(\sqrt{(1-x^2)}}{\sqrt{(1-x^2)^3}}$$ ps

i never used arcsin before and am compeletely unfamilliar with it

Wish
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  • For equations like this, I usually use a computer software like Mathematica. Of course, this will not help you practice though. – David Raveh Jan 09 '23 at 08:38
  • You have some chain rule issues. For example,$$\frac{d}{dx}\frac{1}{\sqrt{1-x^2}}=\frac{x}{\sqrt{1-x^2}^3}.$$Hint: first simplify $\frac{d}{dx}\frac{x}{\sqrt{1-x^2}}$. – J.G. Jan 09 '23 at 08:41
  • @J.G. other than that i have no problems,yes?also i used the rule $n.u(x)^{n-1}.u'x$ rule to get the answer as $(1-x^2)^{3/2}$ note $x^{1/2}$ is same as $\sqrt{x}$ and $\frac{3}{2}$ is $3*\frac{1}{2}$ – Wish Jan 09 '23 at 08:59
  • Of the three terms you should have in your answer, only one matches any of your three terms (update: two matches as of your latest edit). But if you take my hint, you can obtain a simplified expression for the derivative with only two terms. – J.G. Jan 09 '23 at 09:00
  • @J.G. i thought your hint was with my chain rule error and not the actual entire qustion,my fault for not noticing that – Wish Jan 09 '23 at 09:04

4 Answers4

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A faster way is logarithmic differentiation $$y=\frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}}\implies \log(y)=\log(x)+\log(\sin ^{-1}(x))-\frac 12 \log(1-x^2)$$ $$\frac{y'}y=\frac 1 x+\frac{1 } {\sin ^{-1}(x)\sqrt{1-x^2}}+\frac x{1-x^2}=\frac {x \sqrt{1-x^2} +\sin ^{-1}(x)}{x \left(1-x^2\right) \sin ^{-1}(x) }$$

$$y'=\frac{y'}y \times y=\frac {x \sqrt{1-x^2} +\sin ^{-1}(x)}{x \left(1-x^2\right) \sin ^{-1}(x) } \times \frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}}$$ Just simplify

Claude Leibovici
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  1. No.
  2. Since it is false, this is irrelevant.
  3. I would write the function like this $y=x\arcsin x(1-x^2)^{-1/2}$. By triple product rule, $$y'=\arcsin x(1-x^2)^{-1/2}+x\frac{1}{\sqrt{1-x^2}}(1-x^2)^{-1/2}+\\x\arcsin x(-1/2)(1-x^2)^{-3/2}(-2x)$$ $$y'=\frac{\arcsin x}{(1-x^2)^{1/2}}+\frac{x}{{1-x^2}}+\frac{x^2\arcsin x}{(1-x^2)^{3/2}}$$ First and third terms can be combined to give: $$y'=\frac{x}{{1-x^2}}+\frac{\arcsin x}{(1-x^2)^{3/2}}.$$ This is WA.
Bob Dobbs
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You might see that the function can be rewritten as follows:

$$ y=-\arcsin(x)\Big((1-x^2)^{\tfrac{1}{2}}\Big)' $$

Mikasa
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Here a substitution can help out with calculations. Let $ x = \sin t$, i.e. $t = \arcsin x$. First note that $x\in \langle -1,1 \rangle$, so $t\in \langle -\pi/2,\pi/2 \rangle$ and $\cos t>0$. This is important because we can then simplify $\sqrt{\cos^2t} = |\cos t| = \cos t$, which is used couple of times from now on.

Write $$y =\frac{t\sin t}{\sqrt{1-\sin^2t}}=t\tan t. $$ We can now use the chain rule \begin{align}\frac{dy}{dx} &= \frac{dy}{dt} \cdot \frac{dt}{dx}\\ &= (\tan t+\frac{t}{\cos^2t})\frac{dt}{dx} = \left(\frac{x}{\sqrt{1-x^2}}+\frac{\arcsin x}{1-x^2}\right)\frac{1}{\sqrt{1-x^2}}=\frac{x}{1-x^2}+\frac{\arcsin x}{(1-x^2)^{3/2}}\end{align}

where we used $\cos t = \sqrt{1-x^2}$ which is derived directly from $\sin^2t+\cos^2t = 1$.

Ennar
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