Inverse Trigometric Function Question (1 Viewer)

[Rhinoz8142]

y = cos^-1 (cos)

dy/dx =

= -1/√1^2 - (cos x)^2
= -1/√1-cos^2 (x)
= -1/√sin^2 (x)
= (-1)^2 /(√sin^2 (x))^2
= 1/sin^2 (x)


after this I got stuck..

please help

 

[Trebla]

Where did the squares come from in the second last line?

 

[braintic]

You haven't used the chain rule as required.

You should get y=(sinx) / |sinx|.

This is equal to 1 wherever sin x is +ve, -1 when sin x is -ve, and is undefined when sin x is zero.

What was the actual question? Were you asked to differentiate, or were you simply asked to sketch the curve? Because sketching this is best done without calculus.

 

[hit patel]

i think its a sharp point.