Problem 33 - Calculus Single Variable

Question

Calculate the requested derivative. \(f^{\prime \prime \prime}(x)\) where \(f(x)=\cos (4 x+3)\)

Step-by-Step Solution

Answer

The third derivative is \(f'''(x) = 64\sin(4x + 3)\).

Step 1: Differentiate the First Time

To find the first derivative of \(f(x) = \cos(4x + 3)\), use the chain rule. The derivative of \(\cos(u)\) with respect to \(x\) is \(-\sin(u)\) times the derivative of \(u\). Here \(u = 4x + 3\), so \(\frac{du}{dx} = 4\). Thus, the first derivative is \(f'(x) = -\sin(4x + 3) \cdot 4 = -4\sin(4x + 3)\).

Step 2: Differentiate the Second Time

Now, find the second derivative by differentiating \(f'(x) = -4\sin(4x + 3)\). Again, use the chain rule. The derivative of \(\sin(u)\) is \(\cos(u)\), and multiply by \(\frac{du}{dx} = 4\), giving \(f''(x) = -4 \cdot 4\cos(4x + 3) = -16\cos(4x + 3)\).

Step 3: Differentiate the Third Time

Next, differentiate \(f''(x) = -16\cos(4x + 3)\). Using the chain rule, the derivative of \(\cos(u)\) is \(-\sin(u)\), and multiply by \(\frac{du}{dx} = 4\), resulting in \(f'''(x) = -16 \cdot (-4)\sin(4x + 3) = 64\sin(4x + 3)\).

Key Concepts

Understanding the Chain Rule

The chain rule is an essential technique in calculus for differentiating composite functions. A composite function is one that consists of a function within another function. Here, understanding how to apply the chain rule can make calculating derivatives of complex functions more straightforward.
The rule states that to differentiate a composite function, take the derivative of the outer function and multiply it by the derivative of the inner function.

• Suppose you are given a function in the form of \(f(g(x))\), where \(g(x)\) is the inner function and \(f\) is the outer function.
• The derivative \(\frac{d}{dx}(f(g(x)))\) becomes \(f'(g(x)) \cdot g'(x)\).

In our exercise, we see this with \(f(x) = \cos(4x + 3)\). Here, \(u = 4x + 3\) is the inner function (\(g(x)\)), and \(\cos(u)\) is the outer function. Using the chain rule becomes crucial for effectively finding the derivatives.

Finding the First Derivative

The process of finding the first derivative is about understanding how a function changes at any point. In our given problem, the function is \(f(x) = \cos(4x + 3)\). To find the first derivative, we apply the chain rule.

• Differentiate the outer function: The derivative of \(\cos(u)\) with respect to \(x\) is \(-\sin(u)\).
• Multiply by the derivative of the inner function: Here, \(u = 4x + 3\) implies \(\frac{du}{dx} = 4\).
• Thus, \(f'(x) = -\sin(4x + 3) \cdot 4 = -4\sin(4x + 3)\).

With this, you can see how the changes in \(x\) influence the changes in the value of \(f(x)\) through the lens of \(f'(x)\).

Finding the Second Derivative

The second derivative tells us about the concavity of the function and how it curves. After finding the first derivative, \(f'(x) = -4\sin(4x + 3)\), we differentiate again.

• This time, the outer function is \(-4\sin(4x + 3)\).
• The derivative of \(\sin(u)\) is \(\cos(u)\). Don't forget to multiply by \(-4\) due to the coefficient of \(f'(x)\).
• Apply the chain rule: The derivative of the inner function remains the same, \(\frac{du}{dx} = 4\).
• Therefore, \(f''(x) = -4 \cdot 4\cos(4x + 3) = -16\cos(4x + 3)\).

This technique reveals aspects of the function's concavity and can help identify inflection points.

Unveiling the Third Derivative

When calculating the third derivative, we are tapping into how the rate of change of the rate of change evolves. For functions modeling physical scenarios, it might relate to jerk, the rate of change of acceleration. Here, with \(f''(x) = -16\cos(4x + 3)\), let's differentiate once more to find \(f'''(x)\).

• The derivative of \(\cos(u)\) is \(-\sin(u)\). Again, don't forget the constant \(-16\) that comes from \(f''(x)\).
• Apply the chain rule: Multiply by \(-4\), accounting for \(\frac{du}{dx} = 4\).
• This gives \(f'''(x) = -16 \times (-4)\sin(4x + 3) = 64\sin(4x + 3)\).

This third derivative further refines our understanding of how the initial function's behavior evolves across x-values, reflecting the nuances of its changing nature.