Find the derivative of <msqrt> 3 + x </msqrt> e 5 + 4

Use ${\displaystyle \sqrt[n]{a^x}=a^{\frac{x}{n}}}$ to rewrite ${\displaystyle \sqrt{3+x}}$ as ${\displaystyle {(3+x)}^{\frac{1}{2}}}$.

${\displaystyle \frac{d}{dx}\left[{(3+x)}^{\frac{1}{2}}{e}^{5+4x}\right]}$

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 ${\displaystyle f\left(x\right)={(3+x)}^{\frac{1}{2}}}$ and ${\displaystyle g\left(x\right)=e^{5+4x}}$.

${\displaystyle {(3+x)}^{\frac{1}{2}}\frac{d}{dx}\left[e^{5+4x}\right]+e^{5+4x}\frac{d}{dx}\left[{(3+x)}^{\frac{1}{2}}\right]}$

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 ${\displaystyle f\left(x\right)=e^x}$ and ${\displaystyle g\left(x\right)=5+4x}$.

${\displaystyle {(3+x)}^{\frac{1}{2}}\left(e^{5+4x}\frac{d}{dx}[5+4x]\right)+e^{5+4x}\frac{d}{dx}\left[{(3+x)}^{\frac{1}{2}}\right]}$

Differentiate.

${\displaystyle 4\cdot \left({(3+x)}^{\frac{1}{2}}{e}^{5+4x}\right)+e^{5+4x}\frac{d}{dx}\left[{(3+x)}^{\frac{1}{2}}\right]}$

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 ${\displaystyle g\left(x\right)=3+x}$.

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+e^{5+4x}\left(\frac{1}{2}{(3+x)}^{\frac{1}{2}-1}\frac{d}{dx}[3+x]\right)}$

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

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+e^{5+4x}\left(\frac{1}{2}{(3+x)}^{\frac{1}{2}-1\cdot \frac{2}{2}}\frac{d}{dx}[3+x]\right)}$

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

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+e^{5+4x}\left(\frac{1}{2}{(3+x)}^{\frac{1}{2}+\frac{-1\cdot 2}{2}}\frac{d}{dx}[3+x]\right)}$

Combine the numerators over the common denominator.

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+e^{5+4x}\left(\frac{1}{2}{(3+x)}^{\frac{1-1\cdot 2}{2}}\frac{d}{dx}[3+x]\right)}$

Simplify the numerator.

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+e^{5+4x}\left(\frac{1}{2}{(3+x)}^{\frac{-1}{2}}\frac{d}{dx}[3+x]\right)}$

Combine fractions.

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+\frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}\frac{d}{dx}[3+x]}$

By the Sum Rule, the derivative of ${\displaystyle 3+x}$ with respect to ${\displaystyle x}$ is ${\displaystyle \frac{d}{dx}\left[3\right]+\frac{d}{dx}\left[x\right]}$.

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+\frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}(\frac{d}{dx}\left[3\right]+\frac{d}{dx}\left[x\right])}$

Since ${\displaystyle 3}$ is constant with respect to ${\displaystyle x}$, the derivative of ${\displaystyle 3}$ with respect to ${\displaystyle x}$ is ${\displaystyle 0}$.

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+\frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}(0+\frac{d}{dx}\left[x\right])}$

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

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+\frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}\frac{d}{dx}\left[x\right]}$

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

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+\frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}\cdot 1}$

Multiply ${\displaystyle \frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}}$ by 1${\displaystyle 1}$.

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}+\frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}}$

To write ${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}}$ as a fraction with a common denominator, multiply by ${\displaystyle \frac{2{(3+x)}^{\frac{1}{2}}}{2{(3+x)}^{\frac{1}{2}}}}$.

${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}\cdot \frac{2{(3+x)}^{\frac{1}{2}}}{2{(3+x)}^{\frac{1}{2}}}+\frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}}$

Combine ${\displaystyle 4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}}$ and ${\displaystyle \frac{2{(3+x)}^{\frac{1}{2}}}{2{(3+x)}^{\frac{1}{2}}}}$.

${\displaystyle \frac{4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}\left(2{(3+x)}^{\frac{1}{2}}\right)}{2{(3+x)}^{\frac{1}{2}}}+\frac{e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}}$

Combine the numerators over the common denominator.

${\displaystyle \frac{4{(3+x)}^{\frac{1}{2}}{e}^{5+4x}\left(2{(3+x)}^{\frac{1}{2}}\right)+e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}}$

Multiply ${\displaystyle 2}$ by ${\displaystyle 4}$.

${\displaystyle \frac{8{(3+x)}^{\frac{1}{2}}{e}^{5+4x}{(3+x)}^{\frac{1}{2}}+e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}}$

Multiply ${\displaystyle {(3+x)}^{\frac{1}{2}}}$ by ${\displaystyle {(3+x)}^{\frac{1}{2}}}$ by adding the exponents.

${\displaystyle \frac{8{(3+x)}^1{e}^{5+4x}+e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}}$

Simplify ${\displaystyle 8{(3+x)}^1{e}^{5+4x}}$.

${\displaystyle \frac{8(3+x)e^{5+4x}+e^{5+4x}}{2{(3+x)}^{\frac{1}{2}}}}$

Simplify.

$\frac{e^{5+4x}(8x+25)}{2{(3+x)}^{{\displaystyle \frac{1}{2}}}}$