Problem 31 - Calculus and its applications

Question

Differentiate. $$ y=5^{2 x^{3}-1} \cdot \log (6 x+5) $$

Step-by-Step Solution

Answer

The derivative is \(5^{2x^3 - 1} \cdot 6x^2 \ln(5) \cdot \log(6x + 5) + \frac{6 \cdot 5^{2x^3 - 1}}{6x + 5}\).

Step 1: Identify Function Type

The given function is a product of two functions: an exponential function and a logarithmic function. The exponential part is \(5^{2x^3 - 1}\) and the logarithmic part is \(\log(6x + 5)\). We'll need to use the product rule to differentiate this product.

Step 2: Apply the Product Rule

The product rule states that the derivative of a product \(u(x) \cdot v(x)\) is \(u'(x)v(x) + u(x)v'(x)\). Let's identify \(u(x) = 5^{2x^3 - 1}\) and \(v(x) = \log(6x + 5)\). We first need to differentiate each part with respect to \(x\).

Step 3: Differentiate the Exponential Component

To differentiate \(u(x) = 5^{2x^3 - 1}\), use the chain rule. The derivative of \(a^g(x)\) is \(a^g(x) \ln(a) \cdot g'(x)\), where \(g(x) = 2x^3 - 1\). The derivative \(g'(x) = 6x^2\), so the derivative of \(u(x)\) is: \[\frac{d}{dx}(5^{2x^3 - 1}) = 5^{2x^3 - 1} \ln(5) \cdot 6x^2\]

Step 4: Differentiate the Logarithmic Component

Differentiate \(v(x) = \log(6x + 5)\) using the chain rule. The derivative of \(\log(u)\) is \(\frac{1}{u}u'\). Here, \(u = 6x + 5\), so \(u' = 6\). Therefore, \[\frac{d}{dx}(\log(6x + 5)) = \frac{1}{6x + 5} \cdot 6 = \frac{6}{6x + 5}\]

Step 5: Combine Using the Product Rule

Substitute the derivatives from Steps 3 and 4 into the product rule formula:\[\frac{dy}{dx} = (5^{2x^3 - 1} \ln(5) \cdot 6x^2) \cdot \log(6x + 5) + 5^{2x^3 - 1} \cdot \frac{6}{6x + 5}\]

Step 6: Simplify the Expression

Simplify the obtained expression:\[\frac{dy}{dx} = 5^{2x^3 - 1} \cdot 6x^2 \ln(5) \cdot \log(6x + 5) + \frac{6 \cdot 5^{2x^3 - 1}}{6x + 5}\]This is the final expression for the derivative \(\frac{dy}{dx}\).

Key Concepts

Exponential Function

An exponential function in calculus is one of the form \(a^{g(x)}\), where \(a\) is a positive constant base and \(g(x)\) is an exponent that can be any function of \(x\). The exponential function grows rapidly and is widely used in fields like finance, biology, and physics.
The key aspect of exponential functions is their property of constant proportionality, meaning the rate of change (growth or decay) is directly proportional to the value of the function itself.
When differentiating an exponential function using calculus, one must pay attention to both the base and the exponent:

• If the base is \(e\), such as in \(e^x\), the derivative is simply \(e^{g(x)} \cdot g'(x)\).
• If the base is any other positive number, say \(a\), the differentiation requires multiplying by the natural logarithm of the base: \(a^{g(x)} \cdot \ln(a) \cdot g'(x)\).

This involves using another important concept, the chain rule, for general exponents \(g(x)\). Thus, knowing how to handle both the base and exponent is crucial in solving calculus problems involving exponential functions.

Logarithmic Function

Logarithmic functions are the inverse of exponential functions, typically noted as \(\log_b(x)\), where \(b\) is the base. In this context, we usually work with the natural logarithm, where the base \(b\) is \(e\), denoted as \(\ln(x)\).
Logarithms transform multiplicative relationships into additive ones, making them extremely useful for simplifying and analyzing problems involving growth, decay, or multiplicity.
When differentiating a logarithmic function, the derivative is straightforward and employs a combination of basic calculus rules:

• The derivative of \(\ln(x)\) is \(\frac{1}{x}\).
• For functions of the form \(\log(u)\), where \(u\) is a differentiable function of \(x\), the derivative involves applying the chain rule: \(\frac{1}{u} \cdot u'\).

This technique allows us to deal with the more complex relationships found in the logarithmic expressions and ensures accurate results in calculus calculations, especially when combined with other differentiation rules.

Product Rule

The product rule is a fundamental tool in calculus for differentiating the product of two functions. If you have two functions, \(u(x)\) and \(v(x)\), the product rule states that their derivative is:

• \(\frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x)\)

Each term of the result involves taking the derivative of one function while keeping the other function unchanged.
The product rule ensures that you properly consider both components of the product function, capturing how changes in each function affect the product as a whole.
In the context of our original problem, \(u(x)\) is an exponential function, and \(v(x)\) is a logarithmic function. Differentiating each component separately and then combining them according to the product rule gives us the complete derivative. This not only provides accuracy but also transparency in terms of understanding how changes in the individual functions contribute to the overall rate of change.

Chain Rule

The chain rule is pivotal in calculus for differentiating compound functions where a function is nested inside another function. This happens when you have an expression \(f(g(x))\), indicating one function \(f\) of another function \(g\).
The rule can be expressed as:

• \(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

This principle reflects the idea of taking the derivative of the outer function first (while leaving the inner function unchanged), then multiplying by the derivative of the inner function.
For instance, in dealing with the exponential function \(5^{2x^3-1}\) in our problem, the chain rule is used to make sure both the exponent itself \((2x^3 - 1)\) and the base\((5^g(x))\) are accounted for accurately when differentiating. The chain rule is thus indispensable for handling such nested functions efficiently and accurately.