# I’m not following Newton’s derivation of $\frac{d}{dx}(\arcsin x)$

This answer summarizes how Newton determined the derivative of the arcsine function. However, I am having trouble following it to completion. Here is how I understand it thus far (with a bit of narrative in my own words):

Isaac Newton had already extended the binomial theorem to exponents other than nonnegative integers. He took the binomial coefficient to be

$$\binom{n}{r}=\frac{1}{r!} \prod_{i=0}^{r-1}(n-i) = \frac{n(n-1)\cdots(n-r+1)}{r!}$$

and thusly, the generalized binomial theorem

$$(x+y)^n = \sum_{r=0}^\infty \binom{n}{r}x^{n-r}y^r$$

could be applied to expressions such as $\sqrt{1-x^2}$.

This is convenient, because the unit circle in the first and second quadrants is modeled by

\begin{align}y=\sin\theta &= \sqrt{1-x^2} \\ &= \left(1-x^2\right)^{1/2} \\ &= \sum_{r=0}^{\infty} \binom{1/2}{r}(1)^{1/2-r}\left(-x^2\right)^r \\ \end{align}

Computing the initial terms gives

$$\sin\theta = \frac{\left(-x^2\right)^0}{2^0} + \frac{\left(-x^2\right)^1}{2^1} + \frac{\left(-x^2\right)^2}{-2^3} + \frac{\left(-x^2\right)^3}{2^4}+\frac{5\left(-x^2\right)^4}{2^7} + \cdots$$

This is where I get lost. I don’t want to waste time on this method if I’m headed down the wrong path.

Could someone share how it’s done?

It's too long to put in as a comment, so I'll post here instead.

Based on the answer in the link, Newton did not use the method to find the derivative of $\arcsin⁡ x$. He uses the power series of $\arcsin⁡ x$ itself to find the power of series of $\sin⁡ x$ hence $\frac{d}{dx}\sin⁡ x$.

So picking from where you left of, Newton has found that $$\sin x=\sum_{i=0}^\infty{\frac{(-1)^n}{(2n+1)!}x^{2n+1}}$$

After some manipulation, Newton has established that: $$\cos x = \sqrt{1-\sin^2x}\tag{1}$$ and $$\frac{d}{dx}\sin⁡ x=\cos x\tag{2}$$

Now, the usual:

Letting $u=\arcsin x$, we have $$\sin u =x$$ Hence, \begin{align} \frac{du}{dx}&=\frac{du}{d\sin u}\\ &=\left(\frac{d\sin u}{du}\right)^{-1}\\ &=\frac{1}{\cos u}\\ &=\frac{1}{\sqrt{1-\sin^2u}}\\ \frac{d}{dx}\arcsin x&=\frac{1}{\sqrt{1-x^2}}\\ \end{align}

• Of course—the derivative of $\sin$ must be confirmed before one can determine the derivative of $\arcsin$! I feel very silly now Dec 22, 2017 at 17:13
• I feel like this appropriately answers this post. I may end up posting an additional question that requests more detail, but that is no reason not to give you the green check. Dec 22, 2017 at 17:54