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

Question

Aurora Alvarado · Accepted Answer

#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#

Luke Alvarez · Answer

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)#

Christopher Agresta · Answer

undefined To find the derivative of the inverse trigonometric function $ y = \cos(\sec^2(x)) $, we'll need to use the chain rule and the derivative formulas for inverse trig functions.

The chain rule states that if we have a function of a function, we differentiate the outer function first and then multiply by the derivative of the inner function.

The derivative of $ \cos(u) $ is $ -\sin(u) $, and the derivative of $ \sec^2(x) $ is $ 2\sec(x)\tan(x) $.

Applying the chain rule:

$$ \frac{dy}{dx} = -\sin(\sec^2(x)) \times 2\sec(x)\tan(x) $$

This can be simplified as:

$$ \frac{dy}{dx} = -2\sec(x)\tan(x)\sin(\sec^2(x)) $$

So, the derivative of $ y = \cos(\sec^2(x)) $ with respect to $ x $ is $ -2\sec(x)\tan(x)\sin(\sec^2(x)) $.