As a brief referesher, Exponential Functions are commonly expressed as \(f(x) = a^x\), \(a > 0\). They are commonly used to model problems related to Growth and Decay. One of the most common forms of the exponential function is the Natural Exponential Function, \(f(x) = e^x\), where \(e = 2.71828\)..., also referred to as Euler's Constant.
The derivative of an Exponential Function can be expressed as such:
Likewise, the derivative of a Natural Exponential Function can be expressed as such:
As you can see, the derivative of this function is the exact same as the original function. This is because it also takes the product of the Natural Logarithm, \(\text{ln}(e)\). However, since \(\text{ln}(e) = 1\), it simplifies itself to the form of its original function.
Find the derivative of the following exponentials.
i. We can use the Exponential Formula to determine the derivative:
In this equation, we can determine that \(a = 5\). As a result, we can express the derivative as:
\(y' = 5^x\cdot \text{ln}(5)\)
Therefore, we can determine that the derivative of the original equation is \(\boldsymbol{y' = 5^x\cdot \text{ln}5}\).
ii. We can use the Natural Exponential Formula to determine the derivative:
\(\cfrac{d}{dx}\cdot e^x = e^x\)
Since the derivative of a Natural Exponential is the same as the original equation, it can be written as:
\(y' = 4e^x\)
Therefore, we can determine that the derivative of the original equation is \(\boldsymbol{y' = 4e^x}\).
\(y = x^5 \cdot 5^x\)
First, we can identify \(f(x)\) and \(g(x)\) and their respective derivatives.
We can use Power Rule for finding \(f'(x)\):
\(\textcolor{red}{f(x) = x^5}\)
\(\textcolor{red}{f'(x) = 5(x^4)}\)
We can use the Exponential Formula for finding \(g'(x)\):
\(\textcolor{blue}{g(x) = 5^x}\)
\(\textcolor{blue}{g'(x) = 5^x \cdot \text{ln}(5)}\)
Next, we can use Product Rule to determine the derivative of this expression:
\(y' = \textcolor{red}{f'(x)}\textcolor{blue}{g(x)} + \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}\)
\(y' = (\textcolor{red}{5x^4})(\textcolor{blue}{5^x}) + (\textcolor{red}{x^5})(\textcolor{blue}{5^x \cdot \text{ln}(5)})\)
We can simplify further by taking out a common factor (\(5^xx^4\)):
Therefore, we can determine that \(\boldsymbol{y' = 5^xx^4[5 + x \cdot \textbf{ln}(5)]}\).
\(y = \cfrac{2^{4x}}{x^3}\)
First, we can identify \(f(x)\) and \(g(x)\) and their respective derivatives.
We can use Chain Rule and the Exponential Formula for finding \(f'(x)\):
\(\textcolor{red}{f(x) = 2^{4x}}\)
\(\textcolor{red}{f'(x) = 2^{4x} \cdot \text{ln}(2) \cdot 4}\)
We can use Power Rule for finding \(g'(x)\):
\(\textcolor{blue}{g(x) = x^3}\)
\(\textcolor{blue}{g'(x) = 3x^2}\)
We can use Quotient Rule to determine the derivative of this equation:
\(h'(x) = \cfrac{\textcolor{blue}{g(x)}\textcolor{red}{f'(x)} - \textcolor{red}{f(x)}\textcolor{blue}{g'(x)}}{[\textcolor{blue}{g(x)}]^2}\)
\(y' = \cfrac{(\textcolor{blue}{x^3})(\textcolor{red}{2^{4x} \cdot \text{ln}(2) \cdot 4}) - (\textcolor{red}{2^{4x}})(\textcolor{blue}{3x^2})}{(\textcolor{blue}{x^3})^2}\)
\(y' = \cfrac{4x^3\cdot 2^{4x}\cdot \text{ln}(2) - 2^{4x}\cdot 3x^2}{x^6}\)
Next, we can take a common factor (\(x^22^{4x}\)) out of the numerator and simplify to get our final derivative:
\(y' = \cfrac{\cancel{x^2}2^{4x}[4x\cdot \text{ln}(2) - 3]}{x^{\cancel{6}4}}\)
\(y' = \cfrac{2^{4x}[4x\cdot \text{ln}(2) - 3]}{x^4}\)
Therefore, we can determine that \(\boldsymbol{y' = \cfrac{2^{4x}[4x\cdot \textbf{ln}(2) - 3]}{x^4}}\).