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Magazine Chapter 3: Problem 6
Differentiate. $$ g(x)=x^{3}(5.4)^{x} $$
Short Answer
Expert verified
The derivative is \( g'(x) = (5.4)^x (3x^2 + x^3 \ln(5.4)) \).
Step by step solution
01
Recognize the Rule of Differentiation
The function provided is a product of two functions: a polynomial function, \( x^3 \), and an exponential function \( (5.4)^x \). To differentiate this, we'll need to use the product rule, which states that if \( u(x) = f(x) \, g(x) \), then \( u'(x) = f'(x) \, g(x) + f(x) \, g'(x) \). Identify \( f(x) = x^3 \) and \( g(x) = (5.4)^x \).
02
Differentiate the Polynomial Component
Differentiate the polynomial function \( f(x) = x^3 \) using the power rule: \( \frac{d}{dx} x^n = nx^{n-1} \). Applying the formula, we get \( f'(x) = 3x^2 \).
03
Differentiate the Exponential Component
For the exponential function \( g(x) = (5.4)^x \), use the differentiation rule for exponential functions: \( \frac{d}{dx} a^x = a^x \ln(a) \). Therefore, \( g'(x) = (5.4)^x \ln(5.4) \).
04
Apply the Product Rule
Insert the differentiated components back into the product rule equation. So we have: \[ g'(x) = (x^3)'(5.4)^x + x^3((5.4)^x)' \]. Substitute in the derivatives: \[ g'(x) = 3x^2(5.4)^x + x^3(5.4)^x \ln(5.4) \].
05
Simplify the Expression
Combine like terms to simplify the derivative expression: \[ g'(x) = (5.4)^x (3x^2 + x^3 \ln(5.4)) \]. This final expression represents the derivative of the original function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
The product rule is a key tool in calculus typically applied when you need to differentiate a function that is the product of two or more other functions. Imagine you have a function composed of two parts, like in our example where you're dealing with a polynomial and an exponential function multiplied together.
To apply the product rule effectively, here is what you do:
To apply the product rule effectively, here is what you do:
- Identify each part of the product. In our exercise, these are the polynomial part, \( x^3 \) and the exponential part, \( (5.4)^x \).
- Differentiation involves finding the derivative of each part separately first. Once you have both derivatives, you'll then plug them into the product rule formula.
- The formula is: if \( u(x) = f(x) \cdot g(x) \), then the derivative \( u'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x) \).
Polynomial Differentiation
Polynomial differentiation is the process of finding the derivative of a polynomial function. In mathematics, this means following a systematic approach to change the function's power form to its derivative form by using the power rule.
Here鈥檚 a recap of how to differentiate polynomials:
Here鈥檚 a recap of how to differentiate polynomials:
- The power rule states: if \( f(x) = x^n \), the derivative \( f'(x) = nx^{n-1} \).
- For our example, \( x^3 \), applying the power rule involves multiplying the exponent by the coefficient (which is 1 in this basic form) and then subtracting one from the exponent. So, \( f'(x) = 3x^2 \).
Exponential Differentiation
Exponential functions have their own special rules for differentiation. They take the form \( a^x \), where "a" is a constant. The derivative of such a function captures the essence of how exponential growth or decay operates.
Here's the differentiation rule for exponential functions:
Here's the differentiation rule for exponential functions:
- If you have \( g(x) = a^x \), then the derivative \( g'(x) = a^x \ln(a) \).
- Let's apply this rule to our example function, \( (5.4)^x \). The derivative becomes \( (5.4)^x \ln(5.4) \).
