find the derivative of f(x) = arcsin(-x)
Hello, myoplex11!
No logs are involved in this problem . . .
Find the derivative of: $\displaystyle f(x) \:= \:\arcsin(-x)$
As AfterShock pointed out, there is a standard formula for arcsine.
If you are expected to derive this formula, you can do it . . .
We have: .$\displaystyle y \:=\:\arcsin(x)$
Take the sine of both sides: .$\displaystyle \sin y \:=\:x$
Differentiate implicitly: .$\displaystyle \cos y\cdot\frac{dy}{dx} \:=\:1 \quad\Rightarrow\quad \frac{dy}{dx}\:=\:\frac{1}{\cos y}$
Since $\displaystyle \sin y \,= \,\frac{x}{1} \,= \,\frac{opp}{hyp}$, .Pythagorus tell us that: $\displaystyle adj \,= \,\sqrt{1 - x^2}$
. . Hence: $\displaystyle \cos y \,= \,\frac{\sqrt{1 - x^2}}{1} \:=\:\sqrt{1 - x^2}$
And the formula becomes: .$\displaystyle \frac{dy}{dx} \:=\:\frac{1}{\sqrt{1-x^2}} $
The composite function,
$\displaystyle \sin (\sin^{-1} (x))=x$
Now there is a theorem that says that $\displaystyle \sin^{-1}(x)$ is differenciable function so use the chain rule,
$\displaystyle [\sin^{-1}(x)]'\cos (\sin^{-1} (x))=1$
Thus,
$\displaystyle ;\sin^{-1}(x)]'=\frac{1}{\cos (\sin^{-1}) (x)}=\frac{1}{\sqrt{1-x^2}}$
So to find,
$\displaystyle \sin^{-1}(-x)$ derivative use what AfterShock said.