How do you find the derivative of #g(x)=2csc^8(4x)#?

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Answer 1 · 2017-12-20T09:28:33

#dy/dx=-64cot(4x)csc^8(4x)#

Chain rule:
#y=2u^8#
#u=csc(v)#
#v=4x#

#dy/dx=dy/(du)\*(du)/(dv)\*(dv)/(dx)#

#dy/dx=(16u^7)\*(-csc(v)cot(v))\*(4)#

Now we replace every #v# with #4x# and every #u# with #csc(v)\rightarrow csc(4x)#:

#dy/dx=(16csc(4x)^7)\*(-csc(4x)cot(4x))\*(4)#

and rearrange:

#dy/dx=-64csc(4x)cot(4x)csc^7(4x)#

and one more thing to clean up:

#dy/dx=-64cot(4x)csc^8(4x)#