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Answer 1 · 2018-01-03T16:46:42

$\frac{d}{d x} f (x) = \frac{- 3}{2 x^{\frac{3}{2}}} + \frac{1}{\sqrt{x}}$ $d/dxf(x)=(-3)/(2x^(3/2))+1/sqrtx$

$f (x) = \frac{3 + 2 x}{\sqrt{x}}$ $f(x)=(3+2x)/(sqrtx)$

This can also be written as $\to$ $->$

$f (x) = \frac{3}{\sqrt{x}} + \frac{2 x}{\sqrt{x}}$ $f(x)=3/sqrtx+(2x)/sqrtx$

$f (x) = \frac{3}{\sqrt{x}} + \frac{2 \cdot \sqrt{x} \cdot \sqrt{x}}{\sqrt{x}}$ $f(x)=3/sqrtx+(2*sqrtx*sqrtx)/sqrtx$

$f(x)=3/sqrtx+(2cancel(sqrtx)sqrtx)/cancel(sqrtx)$

$f(x)=3/sqrtx+2sqrtx$

Now you can differentiate both sides with respect to $x$

The power rule is $d/dxx^n=nx^(n-1)$

$d/dxf(x)=d/dx3/sqrtx+d/dx2sqrtx$

$d/dxf(x)=(-3)/(2x^(3/2))+1/sqrtx$