Problem 38 - Calculus Early Transcendentals

Question

Suppose that the function \(f\) is differentiable everywhere and \(F(x)=x f(x)\) (a) Express \(F^{\prime \prime \prime}(x)\) in terms of \(x\) and derivatives of \(f\) (b) For \(n \geq 2\), conjecture a formula for \(F^{(n)}(x)\)

Step-by-Step Solution

Answer

\( F'''(x) = x f'''(x) + 3 f''(x) \); \( F^{(n)}(x) = x f^{(n)}(x) + n f^{(n-1)}(x) \) for \( n \geq 2 \).

Step 1: Find the first derivative

To find the first derivative of the function \( F(x) = x f(x) \), we apply the product rule. The product rule states that \((fg)' = f'g + fg'\). Thus, for \( F(x) \), we have:\[ F'(x) = (x f(x))' = x f'(x) + f(x) \]

Step 2: Find the second derivative

Now that we have \( F'(x) = x f'(x) + f(x) \), we differentiate again to find the second derivative. Again, apply the product rule to \( x f'(x) \):\[ F''(x) = (x f'(x) + f(x))' = (x f'(x))' + (f(x))' = x f''(x) + f'(x) + f'(x) = x f''(x) + 2 f'(x) \]

Step 3: Find the third derivative

To find \( F'''(x) \), differentiate \( F''(x) = x f''(x) + 2 f'(x) \):\[ F'''(x) = (x f''(x) + 2 f'(x))' = (x f''(x))' + (2 f'(x))' = x f'''(x) + f''(x) + 2 f''(x) = x f'''(x) + 3 f''(x) \]

Step 4: Conjecture a formula for \(F^{(n)}(x)\)

Observing the pattern from derivatives:- \( F'(x) = x f'(x) + f(x) \)- \( F''(x) = x f''(x) + 2 f'(x) \)- \( F'''(x) = x f'''(x) + 3 f''(x) \)We see the pattern \( F^{(n)}(x) = x f^{(n)}(x) + n f^{(n-1)}(x) \) for \( n \geq 1 \). Thus, we conjecture that for \( n \geq 2 \), the formula is \( F^{(n)}(x) = x f^{(n)}(x) + n f^{(n-1)}(x) \).

Key Concepts

Product Rule

When dealing with differentiation, one of the most essential tools is the product rule. This rule helps us differentiate functions that are products of two or more functions. The product rule states:

• If two functions, say, \(u(x)\) and \(v(x)\), are differentiable, then the derivative of their product is \((uv)' = u'v + uv'\).

This means, when you want to differentiate a product of functions, you do it by taking the derivative of the first function multiplied by the second function, and then add it to the first function multiplied by the derivative of the second function.
Let's apply the product rule to our specific case. For the function \(F(x) = x f(x)\), we see \(u(x) = x\) and \(v(x) = f(x)\). Thus:

• \(u'(x) = 1\) because the derivative of \(x\) is 1.
• \(v'(x) = f'(x)\) because we simply keep the derivative notation for \(f(x)\).

Applying the product rule gives us \(F'(x) = xf'(x) + f(x)\).
This breakdown becomes crucial when differentiating higher-order derivatives as we'll see further.

Higher Order Derivatives

Once you grasp the first derivative using the product rule, you can extend this understanding to higher-order derivatives. A higher-order derivative is simply a derivative of a derivative, and it provides insight into the change of the slope itself. Let's see how these derivatives work.
Starting from our first derivative, \(F'(x) = x f'(x) + f(x)\), we need to apply the product rule again and differentiate each part:

• In \(x f'(x)\), using the product rule, we find the second derivative contribution from it results in \(x f''(x) + f'(x)\).
• For the \(f(x)\) term, differentiating gives \(f'(x)\).

Therefore, combining these results, we have the second derivative: \(F''(x) = x f''(x) + 2 f'(x)\). Here, the coefficient \(2\) in front of \(f'(x)\) comes from summing the derivatives of similar terms.
To get the third derivative, repeat the differentiation process:

• Apply the product rule on \(x f''(x)\) to get \(x f'''(x) + f''(x)\).
• Checking the differentiated \(2 f'(x)\) gives \(2 f''(x)\).

Hence, \(F'''(x) = x f'''(x) + 3 f''(x)\). You will see, as the order of the derivative increases, so does the complexity slightly, but the product rule remains a consistent guide.

Conjecture Formula

Now, armed with the knowledge of how to differentiate using the product rule and derive higher-order derivatives, let's derive and analyze the conjecture formula step by step. Recognizing patterns is key in math. After deriving the first, second, and third derivatives:

• \(F'(x) = x f'(x) + f(x)\)
• \(F''(x) = x f''(x) + 2 f'(x)\)
• \(F'''(x) = x f'''(x) + 3 f''(x)\)

We've observed a consistent pattern where the derivative of order \(n\) seems to follow the formula \(F^{(n)}(x) = x f^{(n)}(x) + n f^{(n-1)}(x)\) for \(n \geq 1\).
This pattern arises from the repetitive application of the product rule and tracking the increase in the coefficients placed in front of derivatives of \(f(x)\).
By continuing with this reasoning, you can make strong conjectures for higher derivatives, ensuring calculations remain manageable by using this reliable formula.