# Question bbdc7

Feb 1, 2017

$= 6 {e}^{2 x}$

#### Explanation:

We will require the use of the chain rule: If $u$ is a function of $x$ and $y$ is a function of $u$, then:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

If we let $u = 2 x$, then:

$\frac{d}{\mathrm{dx}} \left(3 {e}^{2 x}\right) = \frac{d}{\mathrm{du}} \left(3 {e}^{u}\right) \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

$= 3 {e}^{u} \cdot 2 = 6 {e}^{u} = 6 {e}^{2 x}$. (since $\frac{d}{\mathrm{dx}} \left(2 x\right) = 2$)

Note:
If $y$ is a function of $x$, then $\frac{d}{\mathrm{dx}} \left(c y\right) = c \frac{\mathrm{dy}}{\mathrm{dx}}$, where $c$ is a real, nonzero constant.

Feb 1, 2017

$6 {e}^{2 x}$

#### Explanation:

the standard derivative of ${e}^{x} \text{ is } {e}^{x}$

To differentiate ${e}^{2 x} \text{ use the" color(blue)" chain rule}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left({e}^{f \left(x\right)}\right) = {e}^{f \left(x\right)} . f ' \left(x\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow \frac{d}{\mathrm{dx}} \left(3 {e}^{2 x}\right) = 3 {e}^{2 x} . \frac{d}{\mathrm{dx}} \left(2 x\right) = 6 {e}^{2 x}$

Feb 1, 2017

see explanation.

#### Explanation:

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder}}$

$\ln \left(x y\right) = \ln x + \ln y$

rArrln(3e^(2x))≠2xln(3e)#

$\text{but } \ln \left(3 {e}^{2 x}\right) = \ln 3 + \ln {e}^{2 x} = \ln 3 + 2 x \cancel{\ln e}$

$\text{and } \frac{d}{\mathrm{dx}} \left(\ln 3 + 2 x\right) = 2$

$\Rightarrow m ' \left(x\right) = 2 m \left(x\right) = 2.3 {e}^{2 x} = 6 {e}^{2 x}$