# Compute the derivatives of the functions g(x)=(\sqrt{x}+x^{2})(x^{3}+x+\s

Compute the derivatives of the functions
$g\left(x\right)=\left(\sqrt{x}+{x}^{2}\right)\left({x}^{3}+x+\sqrt{2}\right)$
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Step 1
Given
$g\left(x\right)=\left(\sqrt{x}+{x}^{2}\right)\left({x}^{3}+x+\sqrt{2}\right)$
To take derivative of $g\left(x\right)$, we apply product rule
$\left(uv{\right)}^{\prime }={u}^{\prime }v+{v}^{\prime }u$
$g\left(x\right)=\left(\sqrt{x}+{x}^{2}\right)\left({x}^{3}+x+\sqrt{2}\right)$
$u=\sqrt{x}+{x}^{2}$
${u}^{\prime }=\frac{1}{2\sqrt{x}}+2x$
$v={x}^{3}+x+\sqrt{2}$
${v}^{\prime }=3{x}^{2}+1$
Step 2
Now we apply product rule
$g\left(x\right)=\left(\sqrt{x}+{x}^{2}\right)\left({x}^{3}+x+\sqrt{2}\right)$
${g}^{\prime }\left(x\right)=\left(\frac{1}{2\sqrt{x}}+2x\right)\left({x}^{3}+x+\sqrt{2}\right)+\left(3{x}^{2}+1\right)\left(\sqrt{x}+{x}^{2}\right)$

Jeffrey Jordon