Problem 32 - Calculus and its applications

Question

Differentiate. $$ y=\log (7 x+3) \cdot 4^{2 x^{4}+8} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = \frac{7}{7x+3} 4^{2x^4+8} + \log(7x+3) 4^{2x^4+8} \cdot 8x^3 \cdot \ln 4 \).

Step 1: Identify the Structure

The given function is a product of two functions: \[ y = u(x) imes v(x) \] where \( u(x) = \log (7x + 3) \) and \( v(x) = 4^{2x^4 + 8} \). We will differentiate it using the product rule.

Step 2: Recall the Product Rule

The product rule for differentiation states: \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \] We will apply this rule to the function \( y = u(x) imes v(x) \).

Step 3: Differentiate \( u(x) \)

First, we need to differentiate \( u(x) = \log(7x + 3) \). Using the chain rule: \[ u'(x) = \frac{d}{dx}\log(7x + 3) = \frac{1}{7x + 3} \times 7 = \frac{7}{7x + 3} \]

Step 4: Differentiate \( v(x) \)

Now, differentiate \( v(x) = 4^{2x^4 + 8} \). Rewrite \( v(x) \) using \( e \) and natural logarithm:\[ v(x) = e^{(2x^4 + 8) \ln 4} \]Applying the chain rule, we find:\[ v'(x) = \frac{d}{dx}[e^{(2x^4 + 8) \ln 4}] = e^{(2x^4 + 8) \ln 4} \cdot \frac{d}{dx}[2x^4 + 8] \cdot \ln 4 \]Calculating the derivative of \(2x^4 + 8\):\[ \frac{d}{dx}[2x^4 + 8] = 8x^3 \]So,\[ v'(x) = 4^{2x^4 + 8} \cdot 8x^3 \cdot \ln 4 \]

Step 5: Apply the Product Rule

Substitute the derivatives we calculated into the product rule formula:\[ \frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) \]\[ \frac{dy}{dx} = \frac{7}{7x + 3} \cdot 4^{2x^4 + 8} + \log(7x + 3) \cdot 4^{2x^4 + 8} \cdot 8x^3 \cdot \ln 4 \]

Key Concepts

Product Rule

In calculus, the product rule is crucial when you need to differentiate expressions involving products of two or more functions. When you have a function like \[y = u(x) \cdot v(x)\] you apply the product rule to find its derivative.
The product rule formula is: \[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\] In simple terms:

• First, differentiate the first function \(u(x)\) and multiply it by \(v(x)\).
• Then, keep \(u(x)\) as it is and differentiate \(v(x)\).
• Add the two results together.

This rule becomes extremely helpful when the functions themselves are complex, as it simplifies the process significantly.

Chain Rule

The chain rule comes into play when dealing with composite functions, functions inside other functions. It's like peeling an onion—going layer by layer. Take for example a function like \[u(x) = \log(7x + 3)\]. Here, you have an outer function (\(\log\)) and an inner function (\(7x + 3\)).
The chain rule states: \[\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\]In applying this:

• First, differentiate the outer function \(\log(x)\), which becomes \(\frac{1}{x}\), and apply it to the inner function, \(7x + 3\).
• Then, differentiate the inner function, \(7x + 3\), which will give you \(7\).
• Multiply these results.

By following these steps, you break down the differentiation process into smaller, manageable parts.

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions where the usual rules are somewhat cumbersome to apply. It's especially useful when the function is a complicated product, quotient, or involves powers of the variable. In our example, we had the term \[v(x) = 4^{2x^4 + 8}\].
Using the natural logarithm, we can express this as: \[v(x) = e^{(2x^4 + 8) \ln 4}\].Then, the differentiation with respect to \(x\) involves:

• Using the exponential function derivative rule \(e^u\), where \(u\) is a function of \(x\).
• Applying the chain rule to find the derivative of the exponent \(2x^4 + 8\), which results in \(8x^3\).
• Multiplying everything together along with \(\ln 4\).

This method simplifies finding derivatives, especially in cases with exponential and logarithmic functions, allowing us to handle otherwise complex expressions efficiently.