Question #1fad7

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Answer 1 · 2017-10-10T17:09:22

#dy/dx=-1/x^2-1/x^(3/2)#

#y=((sqrtx+1)/sqrtx)^2#

#y=(sqrtx/sqrtx+1/sqrtx)^2#

#y=(1+1/sqrtx)^2#

Using #(a+b)^2=a^2+b^2+2ab#

#y=1^2+1/x+2/sqrtx#

#dy/dx=d/dx(1^2+1/x+2/sqrtx)#

#dy/dx=0+(-1/x^2)+2(-1/(2x^(3/2)))#

#dy/dx=-1/x^2-1/x^(3/2)#