Inverse Trigonometry Function

**Question:**

$\displaystyle f(x) = \sin ^{-1} \left(\frac{x}{a}\right) \implies f'(x) = \frac{1}{\sqrt{a^2-x^2}}$

$\displaystyle f(x) = \cos ^{-1} \left(\frac{x}{a}\right) \implies f'(x) = - \frac{1}{\sqrt{a^2-x^2}}$

These two seem to imply that $\displaystyle \sin ^{-1} x + \cos ^{-1} x= 0$. Why is this not, in fact, being implies? The answer lies in considering constants of integral.

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**Problem Faced:**

I don't see why this isn't the case. For the identity:

$\displaystyle \sin ^2x + \cos ^2x = 0$ differentiated gives $\displaystyle 2\sin x\cos x - 2\sin x \cos x = 0$

So, when $\displaystyle \sin ^{-1} x + \cos ^{-1} x= 1$ is differentiated, it also gives the same result $\displaystyle \left(\frac{1}{\sqrt{a^2-x^2}}\right)$ but change of sign so why is that not a fact? (Worried) Thanks in advance.