Question

How do you find the derivative of `y = cotx(cscx)`?

Answer 1

`(dy)/(dx)=-cscx(csc^2x+cot^2x)`

Explanation:

Product rule states if `f(x)=g(x)h(x)`

then `(df)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)`

Hence as `y=cotx(cscx)`

`(dy)/(dx)=d/(dx)(cotx)xxcscx+d/(dx)(cscx)xxcotx`

= `(-csc^2x)xxcscx+(-cscxcotx)xxcotx`

= `-csc^2x xxcscx-cscxxxcot^2x`

= `-cscx(csc^2x+cot^2x)`