How do you differentiate f(x)=(x-x^1/2)/x^1/9?

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Answer 1 · 2018-02-24T18:01:33

#dy/dx=8/9x^(-1/9)-7/18x^(-11/18)#

Given:

#f(x)=(x-x^(1/2))/(x^(1/9))#

Let #y=f(x)#

Then,

#y=(x-x^(1/2))/(x^(1/9))#

Simplifying further,

#(x-x^(1/2))/(x^(1/9))=(x)/(x^(1/9))-(x^(1/2))/(x^(1/9))#

#y=(x)/(x^(1/9))-(x^(1/2))/(x^(1/9))#

#(x)/(x^(1/9))=x^(1-1/9)=x^(8/9)#

#(x^(1/2))/(x^(1/9))=x^((1/2-1/9))#

#1/2-1/9=9/18-2/18#
#=(9-2)/18=7/18#

#x^((1/2-1/9))=x^(7/18)#

#(x^(1/2))/(x^(1/9))=x^(7/18)#

#(x)/(x^(1/9))-(x^(1/2))/(x^(1/9))=x^(8/9)-x^(7/18)#

#y=x^(8/9)-x^(7/18)#
#(dy)/(dx)=d/(dx)(y)#

#d/dx(y)=d/dx(x^(8/9)-x^(7/18))#

#d/dx(x^(8/9)-x^(7/18))=d/dx(x^(8/9))-d/dx(x^(7/18)))#

#d/dx(x^(8/9))=8/9x^((8/9-1))#

#d/dx(x^(8/9))=8/9x^(-1/9)##

#d/dx(x^(7/18))=7/18x^((7/18-1))#

#d/dx(x^(7/18))=7/18x^(-11/18)#

#d/dx(x^(8/9))-d/dx(x^(7/18))=8/9x^(-1/9)-7/18x^(-11/18)#

#d/dx(x^(8/9)-x^(7/18))=8/9x^(-1/9)-7/18x^(-11/18)#

#d/(dx)(y)=8/9x^(-1/9)-7/18x^(-11/18)#

#dy/dx=8/9x^(-1/9)-7/18x^(-11/18)#