MHB What is the reason for the derivative of arcsin(x) not being -1/sqrt(1-x^2)?

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The derivative of arcsin(x) is derived using the formula for inverse functions, resulting in 1/cos(arcsin(x)), which simplifies to 1/sqrt(1-x^2). The arcsin function is defined with a range of -π/2 to +π/2, ensuring that its derivative remains positive. If the range were altered to allow for a negative derivative, it would no longer represent the arcsin function but rather a different inverse sine function. This distinction clarifies why the derivative cannot be -1/sqrt(1-x^2). Understanding the range of the function is crucial for determining the sign of its derivative.
Petrus
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Hello MHB,
derivate $$\sin^{-1}(x)$$So I use the derivate formula for invers and get
$$\frac{1}{\cos(\sin^{-1}(x))}$$
and Then draw it and get $$\frac{1}{\sqrt{1-x^2}}$$
but there is a reason WHY it can't be $$-\frac{1}{\sqrt{1-x^2}}$$ and I did not understand it, I did not get it.

Regards,
$$|\pi\rangle$$
 
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Re: Derivate of arcsin(x)

Petrus said:
Hello MHB,
derivate $$\sin^{-1}(x)$$So I use the derivate formula for invers and get
$$\frac{1}{\cos(\sin^{-1}(x))}$$
and Then draw it and get $$\frac{1}{\sqrt{1-x^2}}$$
but there is a reason WHY it can't be $$-\frac{1}{\sqrt{1-x^2}}$$ and I did not understand it, I did not get it.

Regards,
$$|\pi\rangle$$

The $\arcsin$ is defined to have a range of $-\pi/2$ to $+\pi/2$.
With this definition the derivative is always positive.
You can also choose the range to be different, making the derivative negative, but then it's not an $\arcsin$ anymore. Then you have a different inverse for the sine.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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