How do you differentiate #y = sqrt(x)(9 x - 8)#?

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Answer 1 · 2016-09-08T06:52:13

#(dy)/(dx)=27/2sqrtx-8/(2sqrtx)#

Product rule states if #f(x)=g(x)h(x)#

then #(df)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)#

Hence as #y=sqrtx(9x-8)#

#(dy)/(dx)=9xxsqrtx+1/(2sqrtx)xx(9x-8)#

= #9sqrtx+(9x)/(2sqrtx)-8/(2sqrtx)#

= #9sqrtx+9/2sqrtx-8/(2sqrtx)#

= #27/2sqrtx-8/(2sqrtx)#