Inverse Trig Question (1 Viewer)

[star*eyed]

QUESTION:
Differentiate tan<SUP>-1</SUP> (x/ square root (1- x<SUP>2</SUP>) )

i think the stuff in the brackets differntiatesto x times sin<SUP>-1</SUP> is <SUP> </SUP>this correct and then do i just use the chain rule???

Anyone with a final answer would be really helpful

 

[sasquatch]

I dunno if you want the working out, so ill put that within a "spoiler tag".

Spoiler

Note: arctan means inverse tan.

Let y =arctan(x/sqrt(1-x²))
Let u = x/sqrt(1-x²)
= x(1-x²)^(-1/2)

y = arctan u

du/dx = (1-x²)^(-1/2) + x * (-1/2) * (1-x²)^(-3/2) * -2x
= 1/(1-x²)^(1/2) + x²/(1-x²)^(3/2)

= 1/(1-x²)^(3/2)


dy/du = 1/(1+u²)
= 1/ ( 1 + {x²/[1-x²]})
= 1-x² (Sorry so hard to write on computer)

dy/dx = du/dx * dy/du
= 1/(1-x²)^(3/2) * (1-x²)
= 1/(1-x²)^(1/2)

Final Answer: dy/dx = 1/(1-x²)^(1/2)

EDIT: Erm... i just realised i never simplified it..sorry its cuz entering on the computer is so crappy..its confusing to look back at stuff cuz of all the sup/sub tags and etc.

 

[star*eyed]

wow! thanks i would NEVER have got that