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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?

Harry Peter
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user88720
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    There's no such thing as "multiplying by sin". You can take the sine of both sides of an equation. – G Tony Jacobs Mar 15 '14 at 05:19
  • From $\cos z\frac{dz}{dx}=1$, we get $\frac{dz}{dx}\frac{1}{\cos z}=\pm \frac{1}{\sqrt{1-\sin^2 z}}$. Since $\sin z=x$ we get $\frac{dz}{dx}=\pm \frac{1}{\sqrt{1-x^2}}$. But $\sin^{-1} x$ is an increasing function on the interval $[-1,1]$. So we choose the $+$, not the $-$. – André Nicolas Mar 15 '14 at 05:23

3 Answers3

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$$\dfrac{dy}{dx}\sin^{-1}(x)^2=\frac{dy}{dx}\left(\sin^{-1}(x)\right)^2$$ 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 $\sin^2(y)=x^2$. $$\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-x^2}}$$ $$\displaystyle \boxed{\therefore \dfrac{dy}{dx}\sin^{-1}(x)=\dfrac{1}{\sqrt{1-x^2}}}$$

Anonymous Computer
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  • Why not just drop the derivatives $dy/dx$ in the first equation? What does it in fact assert? – Ruslan Mar 15 '14 at 09:15
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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 $$\frac{d}{dx}\left[\sin^{-1}(x)\right] = \frac{1}{\sqrt{1-x^2}}.$$

Randy Randerson
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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 $\dfrac{1}{\sqrt{1-x^2}}$.

Imanol Pérez Arribas
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