how to find differential of $\sin^{-1}x$ using first principle?

Question

I am trying to find out the differential of $\sin^{-1}x$ using fist principle as follows
$$\frac{d}{dx}\left(\sin^{-1}x\right)=\lim_{h\to 0}\frac{\sin^{-1}(x+h)-\sin^{-1}(x)}{h}$$ I got stuck here what is next step to reach $\frac{1}{\sqrt{1-x^2}}$. could somebody help me ? thank you very much

Answer 1

$$\frac{d}{dx}\left(\sin^{-1}x\right)=\lim_{h\to 0}\frac{\sin^{-1}(x+h)-\sin^{-1}(x)}{h}$$

Put $\sin^{-1}(x+h)= \alpha \rightarrow x+h=\sin(\alpha)$ and $\sin^{-1}(x)=\beta \rightarrow x=\sin(\beta)$

Therefore $h =\sin \beta-\sin\alpha$

$h \rightarrow0$ implies that $\sin\alpha \rightarrow \sin \beta$

That is $\alpha \rightarrow \beta $ for $-1 \leq \alpha\leq1.$

Then $$\sin^{-1}(x+h)-\sin^{-1}(x)=\alpha-\beta $$

$$\frac{d}{dx}\left(\sin^{-1}x\right)=\lim_{h\to 0}\frac{\sin^{-1}(x+h)-\sin^{-1}(x)}{h}=\lim_{\alpha\to \beta}\left( \frac{\alpha-\beta}{\sin\alpha-\sin \beta}\right)$$

$$\frac{d}{dx}\left(\sin^{-1}x\right)=\lim_{\alpha\to \beta} \frac{\alpha-\beta}{2\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}$$

$$\frac{d}{dx}\left(\sin^{-1}x\right)=\lim_{\alpha\to \beta} \frac{\left(\frac{\alpha-\beta}{2}\right)}{\cos\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)}$$

$$\frac{d}{dx}\left(\sin^{-1}x\right)=\frac{1}{\cos \beta}=\frac{1}{\sqrt{1-x^2}}$$

Answer 2

\begin{align*} \theta &= \sin^{-1} (x+h) \\ \phi &= \sin^{-1} x \\ \sin (\theta-\phi) &= \sin \theta \cos \phi-\sin \phi \cos \theta \\ &= (x+h)\sqrt{1-x^{2}}-x\sqrt{1-(x+h)^{2}} \\ &= \frac{(x+h)^{2}(1-x^{2})-x^{2}[1-(x+h)^{2}]} {(x+h)\sqrt{1-x^{2}}+x\sqrt{1-(x+h)^{2}}} \\ &= \frac{h(2x+h)}{(x+h)\sqrt{1-x^{2}}+x\sqrt{1-(x+h)^{2}}} \\ \lim_{h\to 0} \frac{\sin^{-1}(x+h)-\sin^{-1} x}{h} &= \lim_{h\to 0} \frac{1}{h}\sin^{-1} \left( \frac{2hx}{2x\sqrt{1-x^{2}}} \right) \\ &= \lim_{h\to 0} \frac{1}{h} \sin^{-1} \left( \frac{h}{\sqrt{1-x^{2}}} \right) \\ &= \frac{1}{\sqrt{1-x^{2}}} \lim_{h\to 0} \frac{\sin^{-1} \left( \frac{h}{\sqrt{1-x^{2}}} \right)} {\frac{h}{\sqrt{1-x^{2}}}} \\ &= \frac{1}{\sqrt{1-x^{2}}} \end{align*}