I am shown:
$$f(x) = \arcsin x \implies f'(x) = \frac{1}{\sqrt{1-x^2}}$$
$$f(x) = \arccos x \implies f'(x) = -\frac{1}{\sqrt{1-x^2}}$$
These two derivatives can be very readily derived by a bit of implicit differentiation. However… I want to undo my differentiation for both expressions.
I set up the integrals and now note that the only difference between the two derivatives is a negative sign. Suppose that for the arccos(x) derivative, I factor out the negative sign before integrating the expression, as $-1$ is a constant and the constnat factor rule states that:
$$\int k \frac{dy}{dx} dx = k \int \frac{dy}{dx} dx$$
However, due to this, I end up with the false equality:
$$-\arccos(x)=\arcsin(x)$$
Which doesn't hold for any $x$ at all!
What have I done wrong, why cant one simply factor out the negative sign for the arccos derivative expression?
Best Answer
Two functions defined over an interval that have the same derivative differ by a constant.
Since the derivatives of $\arcsin x$ and $-\arccos x$ are both equal to $\dfrac{1}{\sqrt{1-x^2}}$ on the interval $(-1,1)$, you can say that there exists $c$ such that $$ \arcsin x=c-\arccos x $$ If you evaluate at $0$, you get $$ 0=\arcsin 0=c-\arccos 0=c-\frac{\pi}{2} $$
Therefore, for all $x\in(-1,1)$, $$ \arcsin x=\frac{\pi}{2}-\arccos x $$ and this holds by continuity also for $x=-1$ and $x=1$.
You get a similar “contradiction” if you do $$ \log x=\int\frac{1}{x}\,dx=\int\frac{2}{2x}\,dx=\int\frac{1}{2x}\,d(2x)=\log(2x) $$ without taking into account the “constant of integration”; indeed $$ \log x = \log(2x)-\log 2 $$