# inverse trigonometric differentiation

• Nov 8th 2006, 01:48 PM
myoplex11
inverse trigonometric differentiation
find the derivative of f(x) = arcsin(-x)
• Nov 8th 2006, 01:56 PM
AfterShock
Quote:

Originally Posted by myoplex11
find the derivative of f(x) = arcsin(-x)

The standard derivative is that the derivative of arcsin(x) is:

1/(sqrt(1 - x^2)); it can be proved with some effort.

The derivative of arcsin(-x) then is just -1/(sqrt(1 - x^2)).
• Nov 8th 2006, 02:11 PM
Jameson
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 PM
Soroban
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 PM
ThePerfectHacker
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.