Find the derivative of 1 + x + 7 x 2 </mrow> <mo stretchy="

Use ${\displaystyle \sqrt[n]{a^x}=a^{\frac{x}{n}}}$ to rewrite ${\displaystyle \sqrt[3]{6+4x}}$ as ${\displaystyle {(6+4x)}^{\frac{1}{3}}}$.

${\displaystyle \frac{d}{dx}\left[(1+x+7x^2){(6+4x)}^{\frac{1}{3}}\right]}$

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 ${\displaystyle f\left(x\right)=1+x+7x^2}$ and ${\displaystyle g\left(x\right)={(6+4x)}^{\frac{1}{3}}}$.

${\displaystyle (1+x+7x^2)\frac{d}{dx}\left[{(6+4x)}^{\frac{1}{3}}\right]+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

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 ${\displaystyle f\left(x\right)=x^{\frac{1}{3}}}$ and ${\displaystyle g\left(x\right)=6+4x}$.

${\displaystyle (1+x+7x^2)\left(\frac{1}{3}{(6+4x)}^{\frac{1}{3}-1}\frac{d}{dx}[6+4x]\right)+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

To write ${\displaystyle -1}$ as a fraction with a common denominator, multiply by ${\displaystyle \frac{3}{3}}$.

${\displaystyle (1+x+7x^2)\left(\frac{1}{3}{(6+4x)}^{\frac{1}{3}-1\cdot \frac{3}{3}}\frac{d}{dx}[6+4x]\right)+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

Combine ${\displaystyle -1}$ and ${\displaystyle \frac{3}{3}}$.

${\displaystyle (1+x+7x^2)\left(\frac{1}{3}{(6+4x)}^{\frac{1}{3}+\frac{-1\cdot 3}{3}}\frac{d}{dx}[6+4x]\right)+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

Combine the numerators over the common denominator.

${\displaystyle (1+x+7x^2)\left(\frac{1}{3}{(6+4x)}^{\frac{1-1\cdot 3}{3}}\frac{d}{dx}[6+4x]\right)+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

Simplify the numerator.

${\displaystyle (1+x+7x^2)\left(\frac{1}{3}{(6+4x)}^{\frac{-2}{3}}\frac{d}{dx}[6+4x]\right)+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

Combine fractions.

${\displaystyle (1+x+7x^2)\left(\frac{1}{3{(6+4x)}^{\frac{2}{3}}}\frac{d}{dx}[6+4x]\right)+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

By the Sum Rule, the derivative of ${\displaystyle 6+4x}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[6\right]+\frac{d}{dx}\left[4x\right]}$.

${\displaystyle (1+x+7x^2)\frac{1}{3{(6+4x)}^{\frac{2}{3}}}(\frac{d}{dx}\left[6\right]+\frac{d}{dx}\left[4x\right])+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

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

${\displaystyle (1+x+7x^2)\frac{1}{3{(6+4x)}^{\frac{2}{3}}}(0+\frac{d}{dx}\left[4x\right])+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

Add ${\displaystyle 0}$ and ${\displaystyle \frac{d}{dx}\left[4x\right]}$.

${\displaystyle (1+x+7x^2)\frac{1}{3{(6+4x)}^{\frac{2}{3}}}\frac{d}{dx}\left[4x\right]+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

Since ${\displaystyle 4}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 4x}$ with respect to ${\displaystyle x}$ is ${\displaystyle 4\frac{d}{dx}\left[x\right]}$.

${\displaystyle (1+x+7x^2)\frac{1}{3{(6+4x)}^{\frac{2}{3}}}\left(4\frac{d}{dx}\left[x\right]\right)+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

Combine ${\displaystyle 4}$ and ${\displaystyle \frac{1}{3{(6+4x)}^{\frac{2}{3}}}}$.

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}\frac{d}{dx}\left[x\right]+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

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

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}\cdot 1+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

Multiply ${\displaystyle 1+x+7x^2}$ by ${\displaystyle 1}$.

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}+{(6+4x)}^{\frac{1}{3}}\frac{d}{dx}[1+x+7x^2]}$

By the Sum Rule, the derivative of ${\displaystyle 1+x+7x^2}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[1\right]+\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[7x^2\right]}$.

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}+{(6+4x)}^{\frac{1}{3}}(\frac{d}{dx}\left[1\right]+\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[7x^2\right])}$

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

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}+{(6+4x)}^{\frac{1}{3}}(0+\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[7x^2\right])}$

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

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}+{(6+4x)}^{\frac{1}{3}}(\frac{d}{dx}\left[x\right]+\frac{d}{dx}\left[7x^2\right])}$

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

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}+{(6+4x)}^{\frac{1}{3}}(1+\frac{d}{dx}\left[7x^2\right])}$

Since ${\displaystyle 7}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 7x^2}$ with respect to ${\displaystyle x}$ is ${\displaystyle 7\frac{d}{dx}\left[x^2\right]}$.

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}+{(6+4x)}^{\frac{1}{3}}(1+7\frac{d}{dx}\left[x^2\right])}$

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

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}+{(6+4x)}^{\frac{1}{3}}(1+7\left(2x\right))}$

Multiply ${\displaystyle 2}$ by ${\displaystyle 7}$.

${\displaystyle (1+x+7x^2)\frac{4}{3{(6+4x)}^{\frac{2}{3}}}+{(6+4x)}^{\frac{1}{3}}(1+14x)}$

Simplify.

$\frac{2(98x^2+134x+11)}{3{(6+4x)}^{{\displaystyle \frac{2}{3}}}}$