`y=ln|csc(x)|`
To take the
derivative of this, use the formula:
`(ln u)' = 1/u*u'`
Applying that, y' will be:
`y'=1/(cscx)*(cscx)'`
Take note that the derivative of cosecant is `(csctheta)' = -cscthetacottheta`
.
`y'=1/cscx*(-cscxcotx)`
`y'=-cotx`
Therefore, the derivative of the given function is
`y'=-cotx` .
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