Question

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

Answer 1

`(3*2^(3x))/ln2`

Explanation:

`d/(dx)(2^(3x))`
`=(2^(3x))ln2d/dx(3x)` [suppose a=2 and `d/(dx)(a^x)=a^x/lna`]

`=(3*2^(3x))ln2`

Answer 2

`f'(x)=2^(3x)(3ln(2))`

Explanation:

You may be tempted to differentiate this like it were `e^(3x)`, 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(x)=2^(3x)=e^ln(2^(3x))`

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(a^b)=b*lna`, we know that

`f(x)=e^(3xln(2))`

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

`d/dx(e^u)=e^u*u'`

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

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

`f'(x)=e^(3xln(2))*3ln(2)`

Careful! This is not fully simplified. Recall that `e^(3xln(2))=2^(3x)`, so we can substitute that in.

`f'(x)=2^(3x)(3ln(2))`

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:

`d/dx(a^u)=a^u*ln(a)*u'`

Where `a` is a constant.