find the derivative of f(x) = arcsin(-x)

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- Nov 8th 2006, 01:48 PMmyoplex11inverse trigonometric differentiation
find the derivative of f(x) = arcsin(-x)

- Nov 8th 2006, 01:56 PMAfterShock
- Nov 8th 2006, 02:11 PMJameson
Draw a triangle such that $\displaystyle \sin(y)=x$. Add the hypotenuse and then differentiate both sides of this equation, remembering y is a function of x.

- Nov 8th 2006, 02:16 PMSoroban
Hello, myoplex11!

No logs are involved in this problem . . .

Quote:

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}} $

- Nov 8th 2006, 05:06 PMThePerfectHacker
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.