Find the derivative of 6 + 2 x + x 2 </mrow> <mo stretchy=

Use ${\displaystyle \sqrt[n]{a^x}=a^{\frac{x}{n}}}$ to rewrite ${\displaystyle \sqrt{1+x}}$ as ${\displaystyle {(1+x)}^{\frac{1}{2}}}$.

${\displaystyle \frac{d}{dx}\left[(6+2x+x^2){(1+x)}^{\frac{1}{2}}\right]}$

Differentiate using the Product Rule which states that ${\displaystyle \frac{d}{dx}\left[f\left(x\right)g\left(x\right)\right]}$ is ${\displaystyle f\left(x\right)\frac{d}{dx}\left[g\left(x\right)\right]+g\left(x\right)\frac{d}{dx}\left[f\left(x\right)\right]}$ where ${\displaystyle f\left(x\right)=6+2x+x^2}$ and ${\displaystyle g\left(x\right)={(1+x)}^{\frac{1}{2}}}$.

${\displaystyle (6+2x+x^2)\frac{d}{dx}\left[{(1+x)}^{\frac{1}{2}}\right]+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Differentiate using the chain rule, which states that ${\displaystyle \frac{d}{dx}\left[f\left(g\left(x\right)\right)\right]}$ is ${\displaystyle f\prime \left(g\left(x\right)\right)g\prime \left(x\right)}$ where ${\displaystyle f\left(x\right)=x^{\frac{1}{2}}}$ and ${\displaystyle g\left(x\right)=1+x}$.

${\displaystyle (6+2x+x^2)\left(\frac{1}{2}{(1+x)}^{\frac{1}{2}-1}\frac{d}{dx}[1+x]\right)+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

To write ${\displaystyle -1}$ as a fraction with a common denominator, multiply by ${\displaystyle \frac{2}{2}}$.

${\displaystyle (6+2x+x^2)\left(\frac{1}{2}{(1+x)}^{\frac{1}{2}-1\cdot \frac{2}{2}}\frac{d}{dx}[1+x]\right)+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Combine ${\displaystyle -1}$ and ${\displaystyle \frac{2}{2}}$.

${\displaystyle (6+2x+x^2)\left(\frac{1}{2}{(1+x)}^{\frac{1}{2}+\frac{-1\cdot 2}{2}}\frac{d}{dx}[1+x]\right)+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Combine the numerators over the common denominator.

${\displaystyle (6+2x+x^2)\left(\frac{1}{2}{(1+x)}^{\frac{1-1\cdot 2}{2}}\frac{d}{dx}[1+x]\right)+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Simplify the numerator.

${\displaystyle (6+2x+x^2)\left(\frac{1}{2}{(1+x)}^{\frac{-1}{2}}\frac{d}{dx}[1+x]\right)+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Combine fractions.

${\displaystyle (6+2x+x^2)\left(\frac{1}{2{(1+x)}^{\frac{1}{2}}}\frac{d}{dx}[1+x]\right)+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

By the Sum Rule, the derivative of ${\displaystyle 1+x}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[1\right]+\frac{d}{dx}\left[x\right]}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}(\frac{d}{dx}\left[1\right]+\frac{d}{dx}\left[x\right])+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Since ${\displaystyle 1}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 1}$ with respect to ${\displaystyle x}$ is ${\displaystyle 0}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}(0+\frac{d}{dx}\left[x\right])+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Add ${\displaystyle 0}$ and ${\displaystyle \frac{d}{dx}\left[x\right]}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}\frac{d}{dx}\left[x\right]+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dx}\left[x^n\right]}$ is ${\displaystyle n{x}^{n-1}}$ where ${\displaystyle n=1}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}\cdot 1+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

Multiply ${\displaystyle 6+2x+x^2}$ by ${\displaystyle 1}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}+{(1+x)}^{\frac{1}{2}}\frac{d}{dx}[6+2x+x^2]}$

By the Sum Rule, the derivative of ${\displaystyle 6+2x+x^2}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[6\right]+\frac{d}{dx}\left[2x\right]+\frac{d}{dx}\left[x^2\right]}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}+{(1+x)}^{\frac{1}{2}}(\frac{d}{dx}\left[6\right]+\frac{d}{dx}\left[2x\right]+\frac{d}{dx}\left[x^2\right])}$

Since ${\displaystyle 6}$ is constant with respect to x${\displaystyle x}$, the derivative of ${\displaystyle 6}$ with respect to ${\displaystyle x}$ is ${\displaystyle 0}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}+{(1+x)}^{\frac{1}{2}}(0+\frac{d}{dx}\left[2x\right]+\frac{d}{dx}\left[x^2\right])}$

Add ${\displaystyle 0}$ and ${\displaystyle \frac{d}{dx}\left[2x\right]}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}+{(1+x)}^{\frac{1}{2}}(\frac{d}{dx}\left[2x\right]+\frac{d}{dx}\left[x^2\right])}$

Since ${\displaystyle 2}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 2x}$ with respect to ${\displaystyle x}$ is ${\displaystyle 2\frac{d}{dx}\left[x\right]}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}+{(1+x)}^{\frac{1}{2}}(2\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[x^2\right])}$

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dx}\left[x^n\right]}$ is ${\displaystyle n{x}^{n-1}}$ where ${\displaystyle n=1}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}+{(1+x)}^{\frac{1}{2}}(2\cdot 1+\frac{d}{dx}\left[x^2\right])}$

Multiply 2${\displaystyle 2}$ by 1${\displaystyle 1}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}+{(1+x)}^{\frac{1}{2}}(2+\frac{d}{dx}\left[x^2\right])}$

Differentiate using the Power Rule which states that ${\displaystyle \frac{d}{dx}\left[x^n\right]}$ is ${\displaystyle n{x}^{n-1}}$ where ${\displaystyle n=2}$.

${\displaystyle (6+2x+x^2)\frac{1}{2{(1+x)}^{\frac{1}{2}}}+{(1+x)}^{\frac{1}{2}}(2+2x)}$

Simplify.

$\frac{5(x^2+2x+2)}{2{(1+x)}^{{\displaystyle \frac{1}{2}}}}$