Derivative of the Exponential Function
The derivative of the exponential function ex is $$ D[e^x] = e^x $$. For an exponential function of the form f(x)=ax, where a>0, the derivative is $$ D[a^x] = a^x \cdot \log a $$
A Practical Example
Let’s compute the first derivative of the function:
$$ f(x) = e^{3x} $$
This is a composite function.
So, we need to apply the chain rule for differentiating composite functions.
$$ f'(x) = D[e^{3x}] \cdot D[3x] $$
The derivative of the exponential function is the exponential itself:
$$ f'(x) = e^{3x} \cdot D[3x] $$
The derivative of 3x is simply 3.
$$ f'(x) = e^{3x} \cdot 3 $$
Therefore, the first derivative of the function is:
$$ f'(x) = 3e^{3x} $$
Example 2
Let’s differentiate the following function:
$$ f(x) = a^{2x} $$
This is also a composite function.
Hence, we’ll once again use the chain rule.
$$ f'(x)=D[a^{2x}] \cdot D[2x] $$
First, we differentiate the exponential expression:
$$ f'(x)=a^{2x} \log a \cdot D[2x] $$
Then, differentiating 2x gives us 2.
$$ f'(x)=a^{2x} \log a \cdot 2 $$
So, the derivative of the function simplifies to:
$$ f'(x)= 2 \cdot a^{2x} \log a $$
Proof and Explanation
A] Proof of the Rule $ D[e^x] = e^x $
Consider the exponential function f(x):
$$ f(x)=e^x $$
This function is invertible, and its inverse function is given by:
$$ x = \log f(x) $$
So, we can apply the rule for differentiating inverse functions:
$$ D[f] = \frac{1}{D[f^{-1}]} $$
$$ D[e^x] = \frac{1}{D[\log f(x)]} $$
Knowing that the derivative of the logarithm is 1/x, we get:
$$ D[e^x] = \frac{1}{\frac{1}{f(x)}} $$
$$ D[e^x] = f(x) $$
And since f(x) equals ex:
$$ D[e^x] = e^x $$
Thus, we’ve proven the derivative rule for the exponential function.
B] Proof of the Rule $ D[a^x] = a^x \cdot \log a $
To prove the following derivative rule:
$$ D[a^x] = a^x \cdot \log a $$
we start from the fundamental property of exponential functions:
$$ e^{\log x} = x $$
We can rewrite the derivative in an equivalent form:
$$ D[a^x] $$
$$ D[e^{\log a^x}] $$
$$ D[e^{x \cdot \log a}] $$
Next, we apply the chain rule for composite functions:
$$ D[e^{x \cdot \log a}] \cdot D[ x \log a] $$
$$ e^{x \cdot \log a} \cdot (1 \cdot \log a + x \cdot 0) $$
$$ e^{x \cdot \log a} \cdot \log a $$
$$ e^{\log a^x} \cdot \log a $$
$$ a^x \cdot \log a $$
And that completes the proof of this derivative rule.
And so on.
