Finding derivative of sin^-1 (x^2 + 1)

[thereddevils]

Homework Statement

Diffentiate f(x)=\sin^{-1}(x^2+1)

Homework Equations

The Attempt at a Solution

f'(x)=(-1)(2x)\sin^{-2} (x^2+1)\cos (x^2+1)

am i correct ? But wolfram alpha is giving me something else

http://www.wolframalpha.com/input/?i=differentiate+sin^(-1)+(x^2+1)

 

[tiny-tim]

Hi thereddevils!

(try using the X² tag just above the Reply box )

You're misreading the (admittedly slightly misleading ) special inverse trig notation …

you've correctly differentiated f(x) = 1/sin(x²+1) ,

but the question means f(x) = arcsin(x²+1)

(see eg http://mathworld.wolfram.com/InverseTrigonometricFunctions.html" doesn't help at all )

 

[thereddevils]

thanks tiny , but what's that tag for ? seems that it makes my font smaller .

 

[tiny-tim]

thanks tiny , but what's that tag for ? seems that it makes my font smaller .

Should be smaller, but higher up: 2²² … wheee!

Are you using the same tag as me? …

it's on the second row (the one that starts B I U …), six from the end

 

[thereddevils]

Should be smaller, but higher up: 2²² … wheee!

Are you using the same tag as me? …

it's on the second row (the one that starts B I U …), six from the end

x² , wow never know it can be done that way , interesting !

V_o

Why is there no latex tags in this forum ? I will need to type the tags myself which is sometimes troublesome

 

[tiny-tim]

Why is there no latex tags in this forum ? I will need to type the tags myself which is sometimes troublesome

The ∑ tag (at the end of the line) gives you lots of latex symbols, and the first one you click also gives you [noparse]and[/noparse]

 

[ammontgo]

Table of Derivatives:
[PLAIN]https://dl.dropbox.com/u/4645835/MATH/derv_arcsin.gif

 

[Susanne217]

Remember the Chain rule, reddevils?

which says let

y = y(u(x)) where derivative is

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

which you properly also know as

(f \circ g)'(x) = f'(g(x)) \cdot g'(x)

since you know that

f(u) = sin^{-1}(u)

and

u = x^2+1

then its up to you to use formula above correctly :D

Sincerely
Susanne

 

[thereddevils]

Remember the Chain rule, reddevils?

which says let

y = y(u(x)) where derivative is

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

which you properly also know as

(f \circ g)'(x) = f'(g(x)) \cdot g'(x)

since you know that

f(u) = sin^{-1}(u)

and

u = x^2+1

then its up to you to use formula above correctly :D

Sincerely
Susanne

thanks Susan .

 

[Susanne217]

thanks Susan .

You are welcome!