How do you find the derivative of Inverse trig function #y = cos sec^2(x)#?

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Answer 1 · 2015-08-31T16:51:49

#dy/dx = -2(sin sec^2 x) sec^2xtanx#

As typed, there is no inverse trig function here.

#dy/dxy = -sin(sec^2x) d/dx(sec^2x)#

#= -sin(sec^2x) 2secx d/dx(secx)#

#= -sin(sec^2x) 2secx secx tanx#

# = -2sin (sec^2 x) sec^2xtanx#

Answer 2 · 2015-08-31T23:00:06

I'm guessing the asker meant #"cosecant"#? If so it is written #csc^2x# (although it is not the "inverse trig function" #"arccsc"^2x#).

I will assume he/she actually meant #csc^2x#, the reciprocal trig function of #sinx#, not inverse.

#color(blue)(d/(dx)[csc^2x]) = 2(cscx)*d/(dx)[cscx]#

#= 2cscx*(-cscxcotx)#

#= color(blue)(-2csc^2xcotx)#