# What is the derivative of f(x)= 2^(3x)?

Jan 28, 2016

$\frac{3 \cdot {2}^{3 x}}{\ln} 2$

#### Explanation:

$\frac{d}{\mathrm{dx}} \left({2}^{3 x}\right)$
$= \left({2}^{3 x}\right) \ln 2 \frac{d}{\mathrm{dx}} \left(3 x\right)$ [suppose a=2 and $\frac{d}{\mathrm{dx}} \left({a}^{x}\right) = {a}^{x} / \ln a$]

$= \left(3 \cdot {2}^{3 x}\right) \ln 2$

Jan 28, 2016

$f ' \left(x\right) = {2}^{3 x} \left(3 \ln \left(2\right)\right)$

#### Explanation:

You may be tempted to differentiate this like it were ${e}^{3 x}$, which is very simple to do. However, this cannot be done so easily.

The good thing is, since finding the derivative of an exponential function with base $e$ is so easy, we can make the present function into an exponential function with base $e$.

The first step is recognizing that the following is true:

$f \left(x\right) = {2}^{3 x} = {e}^{\ln} \left({2}^{3 x}\right)$

Since $e$ and $\ln$ are inverses, these functions mean the same thing. Except now, we have a base $e$ so we can differentiate.

Before we differentiate, we can simplify a little more so our differentiation is easier. Through the logarithm rule which states that $\ln \left({a}^{b}\right) = b \cdot \ln a$, we know that

$f \left(x\right) = {e}^{3 x \ln \left(2\right)}$

Now, we can differentiate the function through the chain rule. The chain rule in the case of an exponential $e$ function states that

$\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \cdot u '$

Here, $u = 3 x \ln \left(2\right)$. We know we will have to differentiate $u$, so we might as well do so now. Notice that $u = x \cdot 3 \ln \left(2\right)$, and $3 \ln \left(2\right)$ is just a constant. Thus, $u ' = 3 \ln \left(2\right)$.

Plugging this into the chain rule expression previously identified, we see that

$f ' \left(x\right) = {e}^{3 x \ln \left(2\right)} \cdot 3 \ln \left(2\right)$

Careful! This is not fully simplified. Recall that ${e}^{3 x \ln \left(2\right)} = {2}^{3 x}$, so we can substitute that in.

$f ' \left(x\right) = {2}^{3 x} \left(3 \ln \left(2\right)\right)$

Note that what we just did can be generalized. This is a semi-useful "mold" to commit to memory, although you could do the work every time:

$\frac{d}{\mathrm{dx}} \left({a}^{u}\right) = {a}^{u} \cdot \ln \left(a\right) \cdot u '$

Where $a$ is a constant.