Question #215a5

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Answer 1 · 2017-04-05T09:04:28

The answer is #=2x-1/(2sqrt(x^3))#

We apply the following

#(x^n)'=nx^(n-1)#

#(1/sqrtx)'=(x^(-1/2))'#

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

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

#=-1/(2sqrt(x^3))#

So we have,

#y=x^2+3/sqrtx#

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

I hope that this is a little clearer for you.