differentiating inverse trig (1 Viewer)

[zinc]

how do you differentiate
y=tan-1(1/x) and y=1/(sin-1x)
thanks

 

[independantz]

y=arctan(1/x)
y'=1/(1+(1/x)^2)


y=1/arcsinx=(arcsinx)^-1
y'=(-1)(arcsinx)^-2 x(1/sqrt(1-x^2)), by the chain rule.
=-1/(sqrt(1-x^2))(arcsinx)^2

 

[zinc]

i see, thanks

 

[vds700]

y=arctan(1/x)
y'=1/(1+(1/x)^2)


y=1/arcsinx=(arcsinx)^-1
y'=(-1)(arcsinx)^-2 x(1/sqrt(1-x^2)), by the chain rule.
=-1/(sqrt(1-x^2))(arcsinx)^2

the first one is wrong...

y = arctan(1/x) , let u = 1/x = x^-1
dy/dx = dy/du . du/dx
=1/(1 + u^2). -x^-2
= -1/{[1 + (1/x^2)] . x^2}
= -1/(x^2 + 1)

 

[lyounamu]

the first one is wrong...

y = arctan(1/x) , let u = 1/x = x^-1
dy/dx = dy/du . du/dx
=1/(1 + u^2). -x^-2
= -1/{[1 + (1/x^2)] . x^2}
= -1/(x^2 + 1)

Yeah, I think Andrew is right based on his working out.