Problem 3
Question
Differentiate. $$y=\log _{b} x^{3}$$
Step-by-Step Solution
Verified Answer
\(y' = \frac{3}{x \ln b}\)
1Step 1: Apply the logarithm power rule
Use the logarithm power rule, which states that \(\log_{b}(x^n) = n \cdot \log_{b}(x)\), to move the exponent in front of the logarithm. In this case, we can rewrite the given function as \(\ y = 3 \cdot \log_{b} x\).
2Step 2: Differentiate the function
Use the derivative of a logarithmic function, which is \(\frac{d}{dx} \log_{b} x = \frac{1}{x \ln b}\), to differentiate \(\ y = 3 \cdot \log_{b} x\). Applying the constant multiple rule, the derivative of \(\ y\) with respect to \(\ x\) is \(\ y' = 3 \cdot \frac{1}{x \ln b}\).
3Step 3: Simplify the derivative
Simplify the expression to obtain the final derivative of the function, which is \(\ y' = \frac{3}{x \ln b}\).
Key Concepts
Logarithm Power RuleDerivative of Logarithmic FunctionConstant Multiple Rule
Logarithm Power Rule
Understanding the logarithm power rule is a critical step in differentiating logarithmic functions. In essence, this rule allows us to simplify logarithmic expressions that involve exponents by moving the exponent in front of the logarithm. The general form of this rule is expressed as \( \log_{b}(x^n) = n \cdot \log_{b}(x) \).
For instance, if you have a function like \( y=\log _{b} x^{3} \), using the logarithm power rule would transform it into \( y = 3 \cdot \log_{b} x \). This not only simplifies the expression but also prepares it for differentiation by making the exponent a coefficient, which is much easier to handle when applying derivatives.
For instance, if you have a function like \( y=\log _{b} x^{3} \), using the logarithm power rule would transform it into \( y = 3 \cdot \log_{b} x \). This not only simplifies the expression but also prepares it for differentiation by making the exponent a coefficient, which is much easier to handle when applying derivatives.
Derivative of Logarithmic Function
When it comes to the derivative of a logarithmic function, the process is about understanding how the rate of change of a logarithmic function behaves. For the base \( b \) logarithm of \( x \), the derivative is given by the formula \( \frac{d}{dx} \log_{b} x = \frac{1}{x \ln b} \), where \( \ln \) denotes the natural logarithm.
As seen in our example after applying the logarithm power rule, the next step is to differentiate the resulting function \( y = 3 \cdot \log_{b} x \). Utilizing the derivative formula, we get \( y' = 3 \cdot \frac{1}{x \ln b} \), displaying how each part of the logarithm contributes to the rate at which \( y \) changes with respect to \( x \).
As seen in our example after applying the logarithm power rule, the next step is to differentiate the resulting function \( y = 3 \cdot \log_{b} x \). Utilizing the derivative formula, we get \( y' = 3 \cdot \frac{1}{x \ln b} \), displaying how each part of the logarithm contributes to the rate at which \( y \) changes with respect to \( x \).
Constant Multiple Rule
The constant multiple rule in calculus is all about efficiency. It tells us that if you multiply a function by a constant, its derivative is simply the constant multiplied by the derivative of the function. Mathematically, it's represented as \( \frac{d}{dx}[cf(x)] = c\frac{d}{dx}f(x) \), where \( c \) is a constant.
Referring back to our function \( y = 3 \cdot \log_{b} x \), applying the constant multiple rule to its derivative means you would multiply the constant \( 3 \) by the derivative of \( \log_{b} x \), resulting in \( y' = 3 \cdot \frac{1}{x \ln b} \) as we've seen in our example. This rule helps streamline the differentiation process and leads to a final, simplified expression of the derivative.
Referring back to our function \( y = 3 \cdot \log_{b} x \), applying the constant multiple rule to its derivative means you would multiply the constant \( 3 \) by the derivative of \( \log_{b} x \), resulting in \( y' = 3 \cdot \frac{1}{x \ln b} \) as we've seen in our example. This rule helps streamline the differentiation process and leads to a final, simplified expression of the derivative.