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

2 Answers
Jan 28, 2016

Answer:

#(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#

Jan 28, 2016

Answer:

#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.