How do you find the derivative of #[cos(2x^4) - 1]/x^7#?

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Answer 1 · 2018-03-21T13:45:41

#dy/dx=-8/x^4cosx^4+14/x^8sin^2x^4#

Let
#y=(cos(2x^4)-1)/(x^7)#

Let
# u=cos(2x^4)-1#
#v=x^7#

#y=u/v#

#dy/dx=(v(du)/dx-u(dv)/dx)/v^2#

# u=cos(2x^4)-1#
Let
#t=x^4#

#u=cos2t-1#
wkt
#1-cos2t=2sin^2t#
#u=-(1-cos2t)#
#u=-2sin^2t#
By chain rule
#(du)/dx=(du)/dt(dt)/dx#

#u=-2sin^2t#
#sin^2t=(sint)^2#
Let
#p=sint#

#(sint)^2=p^2#

#u=-2p#
By chain rule
#(du)/dt=-2(dp)/dt#

#p=sint#

#(dp)/dt=cost#

#(du)/dt=-2cost#

#-2cost=-2cosx^4#

#(du)/dt=-2cosx^4#

#(du)/dx=(du)/dt(dt)/dx#

#t=x^4#

#(dt)/dx=4x^3#

#(du)/dx=-2cosx^4(4x^3)#

#(du)/dx=-8x^3cosx^4#

#v=x^7#

#(dv)/dx=7x^6#

#dy/dx=(v(du)/dx-u(dv)/dx)/v^2#

# u=-2sin^2t#
#u=-2sin^2x^4#
#v=x^7#
#(du)/dx=-8x^3cosx^4#
#(dv)/dx=7x^6#

#dy/dx=(x^7(-8x^3cosx^4)-(-2sin^2x^4)(7x^6))/(x^7)^2#

#dy/dx=(-8x^10cosx^4+14x^6sin^2x^4)/x^14#

#dy/dx=-8/x^4cosx^4+14/x^8sin^2x^4#