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 $y=sqrtx(9x-8)$

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

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

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

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

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