# Second derivative of $e^{2 x}$

The calculator will find the second derivative of $e^{2 x}$, with steps shown.

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Find $\frac{d^{2}}{dx^{2}} \left(e^{2 x}\right)$.

### Find the first derivative $\frac{d}{dx} \left(e^{2 x}\right)$

The function $e^{2 x}$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = e^{u}$ and $g{\left(x \right)} = 2 x$.

Apply the chain rule $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$:

$${\color{red}\left(\frac{d}{dx} \left(e^{2 x}\right)\right)} = {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(2 x\right)\right)}$$

The derivative of the exponential is $\frac{d}{du} \left(e^{u}\right) = e^{u}$:

$${\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(2 x\right) = {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(2 x\right)$$

$$e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(2 x\right) = e^{{\color{red}\left(2 x\right)}} \frac{d}{dx} \left(2 x\right)$$

Apply the constant multiple rule $\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$ with $c = 2$ and $f{\left(x \right)} = x$:

$$e^{2 x} {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} = e^{2 x} {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)}$$

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:

$$2 e^{2 x} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 2 e^{2 x} {\color{red}\left(1\right)}$$

Thus, $\frac{d}{dx} \left(e^{2 x}\right) = 2 e^{2 x}$.

### Next, $\frac{d^{2}}{dx^{2}} \left(e^{2 x}\right) = \frac{d}{dx} \left(2 e^{2 x}\right)$

Apply the constant multiple rule $\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$ with $c = 2$ and $f{\left(x \right)} = e^{2 x}$:

$${\color{red}\left(\frac{d}{dx} \left(2 e^{2 x}\right)\right)} = {\color{red}\left(2 \frac{d}{dx} \left(e^{2 x}\right)\right)}$$

The function $e^{2 x}$ is the composition $f{\left(g{\left(x \right)} \right)}$ of two functions $f{\left(u \right)} = e^{u}$ and $g{\left(x \right)} = 2 x$.

Apply the chain rule $\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$:

$$2 {\color{red}\left(\frac{d}{dx} \left(e^{2 x}\right)\right)} = 2 {\color{red}\left(\frac{d}{du} \left(e^{u}\right) \frac{d}{dx} \left(2 x\right)\right)}$$

The derivative of the exponential is $\frac{d}{du} \left(e^{u}\right) = e^{u}$:

$$2 {\color{red}\left(\frac{d}{du} \left(e^{u}\right)\right)} \frac{d}{dx} \left(2 x\right) = 2 {\color{red}\left(e^{u}\right)} \frac{d}{dx} \left(2 x\right)$$

$$2 e^{{\color{red}\left(u\right)}} \frac{d}{dx} \left(2 x\right) = 2 e^{{\color{red}\left(2 x\right)}} \frac{d}{dx} \left(2 x\right)$$

Apply the constant multiple rule $\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$ with $c = 2$ and $f{\left(x \right)} = x$:

$$2 e^{2 x} {\color{red}\left(\frac{d}{dx} \left(2 x\right)\right)} = 2 e^{2 x} {\color{red}\left(2 \frac{d}{dx} \left(x\right)\right)}$$

Apply the power rule $\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$ with $n = 1$, in other words, $\frac{d}{dx} \left(x\right) = 1$:

$$4 e^{2 x} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)} = 4 e^{2 x} {\color{red}\left(1\right)}$$

Thus, $\frac{d}{dx} \left(2 e^{2 x}\right) = 4 e^{2 x}$.

Therefore, $\frac{d^{2}}{dx^{2}} \left(e^{2 x}\right) = 4 e^{2 x}$.

$\frac{d^{2}}{dx^{2}} \left(e^{2 x}\right) = 4 e^{2 x}$A