Derivatives of inverse trig functions (1 Viewer)

[cutemouse]

Hello,

For the derivatives of inverse trig functions (eg. d/dx sin^-1(1-x)), can we just use the results, or do we have to derive it by going through the chain rules and stuff?

Thanks

 

[Absolutezero]

If memory serves me correctly, you should just be able to to use the result.

 

[Timothy.Siu]

u can use results

 

[Arowana21]

u wanna derive it. that is 4 unit stuff from what i can remember

 

[Timothy.Siu]

u wanna derive it. that is 4 unit stuff from what i can remember

nah its 3unit, integrating it is 4unit

 

[cutemouse]

No, this is Ext 1 stuff.

 

[kevinant]

It's 3U stuffs
sin y=x
dx/dy=cos x
dy/dx=1/cos x
=1/sqrt(1-x^2)
d/dx sin-1(1-x)
=-1/sqrt[1-(1-x)^2]
*use chain rule.

 

[cutemouse]

Hello again,

For the domain and range for the derivatives of inverse trig functions, I've noticed that it pretty much is the range of the original inverse function, but without the 'equals' bit in the inequality. ie, if the domain of the original trig function was -1 _< x _< 1 then the derivatives' domain is -1 < x < 1

So is this always the case, or do I need to work it out?

Thanks

 

[Timothy.Siu]

Hello again,

For the domain and range for the derivatives of inverse trig functions, I've noticed that it pretty much is the range of the original inverse function, but without the 'equals' bit in the inequality. ie, if the domain of the original trig function was -1 _< x _< 1 then the derivatives' domain is -1 < x < 1

So is this always the case, or do I need to work it out?

Thanks

because u cant differentiate the end of a line, theres no tangent there...basically

 

[cutemouse]

So is that always the case (with the domain), or can it differ?

 

[Trebla]

The derivative of sin^-1x and cos^-1x (excludes tan^-1x) is UNDEFINED at |x| = 1, as in theory a tangent to those points would be virtually vertical.

 

[Timothy.Siu]

also, umm theres another reason...

y=sin^-1x
y'=1/sqrt(1-x²)

y' would be undefined if x=+-1
but its because yeah u just cant different the end of a line.

 

[cutemouse]

Yeah... so can I always 'assume' that if the domain of the derivative exclude the equals equality sign for the given function, but is the same values?

 

[Timothy.Siu]

Yeah... so can I always 'assume' that if the domain of the derivative exclude the equals equality sign for the given function, but is the same values?

yes for inverse sin and cos

 

[cutemouse]

Hello,

I have one more question.

Find the derivative of tan(sin^-1x)

Thanks for all those helping me.

 

[Trebla]

Hello,

I have one more question.

Find the derivative of tan(sin^-1x)

Thanks for all those helping me.

d(tan(sin^-1x)/dx = sec²(sin^-1x) / √(1 - x²)

 

[cutemouse]

Thanks for that... Just wondering, how do you get the denominator?

Thanks

 

[Timothy.Siu]

Thanks for that... Just wondering, how do you get the denominator?

Thanks

seems like derivative of sin^-1x from chain rule or sometihng
actually it is.