Answer
$f′(x)=\frac{(3x^{2}-2)}{2\sqrt {x^{3}-2x+5}}$
Work Step by Step
Given that $f(x)=\sqrt {x^{3}-2x+5}$
First write it as $f(x)=({x^{3}-2x+5})^{\frac{1}{2}}$ so that it is easier to deal with.
Next we write this as $f(x)=(g(x))^\frac{1}{2}$, where $f(x)$ is the outside function and $g(x)={x^{3}-2x+5}$ is the inside function.
To find the answer we need to differentiate the outside function $f(x)$ and times it by the derivative of the inside function $g(x)$.
$f(x)=({x^{3}-2x+5})^{\frac{1}{2}}$
$f(x)=(g(x))^\frac{1}{2}$
$f′(x)=\frac{1}{2}(g(x))^{\frac{1}{2}-1}×(g′(x))$
$f′(x)=\frac{1}{2}(g(x))^{\frac{-1}{2}}×(3x^{2}-2)$
$f′(x)=\frac{1}{2}({x^{3}-2x+5})^{\frac{-1}{2}}×(3x^{2}-2)$
$f′(x)=\frac{1}{2}(3x^{2}-2)({x^{3}-2x+5})^{\frac{-1}{2}}$
$f′(x)=\frac{(3x^{2}-2)}{2\sqrt {x^{3}-2x+5}}$
