What is the derivative of #sec^-3(x)#?

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Answer 1 · 2018-07-26T19:25:33

#(dy)/(dx)=-3sinxcos^2x#

Here ,

#y=sec^(-3)(x)=1/sec^3x=cos^3x#

#"Using"color(blue)" Chain Rule :"#

#(dy)/(dx)=3cos^2xd/(dx)(cosx)#

#:.(dy)/(dx)=3cos^2x(-sinx)=-3sinxcos^2x#

Answer 2 · 2018-07-26T19:27:39

#1/3\cos^3(\sec^{-3}x)\cot (sec^{-3}x)#

Given function:

#y=\sec^{-3}(x)#

#x=\sec^3y#

Now, differentiating above equation w.r.t. on both the sides by using chain rule as follows

#d/dx(x)=d/dx(\sec^3y)#

#1=3\sec^2yd/dx(\sec y)#

#1=3\sec^2y(sec y\tan y)dy/dx#

#1=3\sec^3y\tan ydy/dx#

#dy/dx=1/3\cos^3y\cot y#

#dy/dx=1/3\cos^3(\sec^{-3}x)\cot (sec^{-3}x)#

Answer 3 · 2018-07-26T19:27:46

#-3cos^2(x)sin(x)#

We have #sec(x)=(cos(x))^(-1)# so #(sec(x))^(-3)=cos(x)^3#
so we get by the chain rule

#(cos(x)^3)'=3cos(x)^2(-sin(x))=-3cos^2(x)sin(x)#