So $y=\sin^{-1}(x)^2$
I am asked to find $\frac{dy}{dx}$
Using the chain rule I find
$\frac{dy}{dx}$= $2\sin^{-1}(x) * \frac{d}{dx}(\sin^{-1}(x))$
I let
$z = \sin^{-1}(x)$
Multiplying both sides by sin
$\sin(z)=\sin(\sin^{-1}(x))$
$\sin(z)=x$
I want to find the derivative of z so I differentiate both sides
$\frac{d}{dx}(\sin(z))=\frac{dx}{dx}$ = 1
Using chain rule I find $\frac{d}{dx}(\sin(z))=$
$\cos(z)\frac{dz}{dx}=1$
So I move $\cos(z)$ over
$\frac{dz}{dx} = \frac{1}{\cos(z)}$
I know that $1 = \cos^2(z)+\sin^2(z)$ so
$\cos^2(z)=1-\sin^2(z)$
$\cos(z)=\frac{+}{-}\sqrt{1-\sin^2(z)}$
And this is where I am lost. The working says that from there, $\cos(z)$ is somehow
$=\frac{1}{\sqrt{1-x^2}}$
I am not sure how they came to that. Can someone please explain how they made that leap?
$\endgroup$ 24 Answers
$\begingroup$$$\dfrac{dy}{dx}\sin^{-1}(x)^2=\frac{dy}{dx}\left(\sin^{-1}(x)\right)^2$$ It can be seen that this is a composition of two functions $f(g(x))$, where $f(x)=x^2$ and $g(x)=\sin^{-1}(x)$. Therefore we need to apply chain rule to this. The chain rule is: $$(f\circ g)'(x)= f'(g(x))\cdot g(x)$$ Let,s apply that to our derivative. $$\dfrac{dy}{dx}\left(\sin^{-1}(x))\right)^2$$ $$=2\left(\sin^{-1}(x)\right)^1\cdot \dfrac{dy}{dx}\sin^{-1}(x)$$ There is a well-known derivative for $\sin^{-1}(x)$. It is $\dfrac{1}{\sqrt{1-x^2}}$. $$=2\sin^{-1}(x)\cdot \dfrac{1}{\sqrt{1-x^2}}$$ $$=\dfrac{2\sin^{-1}(x)}{\sqrt{1-x^2}}$$ $$\displaystyle \boxed{\therefore \dfrac{dy}{dx}\sin^{-1}(x)^2=\dfrac{2\sin^{-1}(x)}{\sqrt{1-x^2}}}$$
DERIVATION OF $\dfrac{dy}{dx}\sin^{-1}(x)$:
Want to know how to derive $\dfrac{dy}{dx}\sin^{-1}(x)$? I will show you. It can be found using implicit differentiation. $$y=\sin^{-1}(x), \ \text{so} \ x=\sin(y)$$ We also know that: $$-\dfrac{\pi}{2}\le y \le \dfrac{\pi}{2}$$Proof:$$\dfrac{d}{dx}x=\dfrac{d}{dx}\sin(y)$$ $$1=\dfrac{d}{dy}\sin(y)\dfrac{dy}{dx}$$ $$1=\cos(y)\dfrac{dy}{dx}$$ $$\dfrac{dy}{dx}=\dfrac{1}{\cos(y)}$$ Remember the Pythagorean identity: $$\sin^2(y)+\cos^2(y)=1$$ $$\cos^2(y)=1-\sin^2(y)$$ $$\cos(y)=\pm\sqrt{1-\sin^2(y)}$$ Because $\cos(y)$ has to be positive (remember, $-\frac{\pi}{2}\leq y \leq \frac{\pi}{2}$), $\cos(y)=\sqrt{1-\sin^2(y)}$. $$\dfrac{dy}{dx}=\dfrac{1}{\cos(y)}=\dfrac{1}{\sqrt{1-\sin^2(y)}}$$ Remember that $\sin(y)=x$. So $\sin^2(y)=x^2$. $$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$$ $$\displaystyle \boxed{\therefore \dfrac{dy}{dx}\sin^{-1}(x)=\dfrac{1}{\sqrt{1-x^2}}}$$
$\endgroup$ 1 $\begingroup$I think they mean $$\frac{dz}{dx} = \frac{1}{\sqrt{1-x^2}},$$ since $$\frac{dz}{dx} = \frac{1}{\cos(z)}$$ and $$x = \sin(z) \implies \cos(z) = \sqrt{1-\sin^2(z)} = \sqrt{1 - x^2}.$$ There is also a rule that $$\frac{d}{dx}\left[\sin^{-1}(x)\right] = \frac{1}{\sqrt{1-x^2}}.$$
$\endgroup$ 4 $\begingroup$If $y=\sin^{-1} (x)^2$, as you have pointed out:
$$\dfrac{dy}{dx} = 2\sin^{-1}(x) \cdot \frac{d}{dx}(\sin^{-1}(x))$$
We want to find now $\frac{d}{dx}(\sin^{-1}(x))$.
Let $f:[-\frac{\pi}{2}, \frac{\pi}{2}] \longrightarrow [0, 1]$ be $f(x) = \sin(x)$. Since $f\circ f^{-1} = id$
Therefore:
$$(f \circ f^{-1})' = (id)'$$
By the chain rule:
$$f' \circ f^{-1} \cdot (f^{-1})' = 1$$
$$(f^{-1})' = \dfrac{1}{f' \circ f^{-1}}$$
Therefore:
$$\dfrac{d}{dx} \sin ^{-1} (x) = \dfrac{1}{\cos (\sin^{-1} (x))}$$
Which, using trigonometric identities, is equal to $\dfrac{1}{\sqrt{1-x^2}}$.
$\endgroup$ $\begingroup$The series expansion\begin{equation} \biggl(\frac{\arcsin x}{x}\biggr)^2=2!\sum_{k=0}^{\infty} [(2k)!!]^2 \frac{x^{2k}}{(2k+2)!}, \quad |x|<1 \end{equation}can be reformulated as\begin{equation} (\arcsin x)^2=2!\sum_{k=0}^{\infty} [(2k)!!]^2 \frac{x^{2k+2}}{(2k+2)!}, \quad |x|<1. \end{equation}Differentiating on both sides of the above equality gives\begin{equation} \bigl[(\arcsin x)^2\bigr]'=2!\sum_{k=0}^{\infty} [(2k)!!]^2 \frac{x^{2k+1}}{(2k+1)!} =\frac{2\arcsin x}{\sqrt{1-x^2}\,}, \quad |x|<1. \end{equation}
Reference
- Bai-Ni Guo, Dongkyu Lim, and Feng Qi, Maclaurin series expansions for powers of inverse (hyperbolic) sine, for powers of inverse (hyperbolic) tangent, and for incomplete gamma functions, with applications, arXiv (2021), available online at .
