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.