Question

Consider the function \(f(x)=g(x) h(x)\). (a) Use the Product Rule to generate rules for finding \(f^{\prime \prime}(x)\), \(f^{\prime \prime \prime}(x),\) and \(f^{(4)}(x)\) (b) Use the results in part (a) to write a general rule for \(f^{(n)}(x)\).

Short answer

The nth derivative of the function \( f(x) = g(x)h(x) \) is given by the sum of terms, where each term is a derivative of \(g(x)\) times a derivative of \(h(x)\), and the exponents add to \(n\). The binomial coefficient in each term indicates the number of ways to obtain that particular product of derivatives. The formula is \(f^{(n)}(x) = ∑_{k=0}^{n} C(n, k) g^{(n-k)}(x) h^{(k)}(x)\), where \(C(n, k)\) is the binomial coefficient 'n choose k'.

Step-by-step solution

Apply the Product Rule to find \( f' (x) \)

Using the product rule which states that the derivative of the product of two functions is given by the first function times the derivative of the second function plus the second function times the derivative of the first function, we have \( f'(x) = g'(x)h(x) + g(x)h'(x) \)

Apply the Product Rule to find \( f''(x) \)

Now, we find the second derivative by differentiating \( f'(x) = g'(x)h(x) + g(x)h'(x) \). Here, each term is a product, so we apply the product rule to both terms individually: \( f''(x) = g''(x)h(x) + 2g'(x)h'(x) + g(x)h''(x) \)

Apply the Product Rule to find \(f'''(x)\)

Applying the same process used for \( f''(x) \), we get \( f'''(x) = g'''(x)h(x) + 3g''(x)h'(x) + 3g'(x)h''(x) + g(x)h'''(x) \)

Apply the Product Rule to find \(f^{(4)}(x)\)

Applying the same procedure for \(f^{(4)}(x)\) we get \( f^{(4)}(x) = g^{(4)}(x)h(x) + 4g'''(x)h'(x) + 6g''(x)h''(x) + 4g'(x)h'''(x) + g(x)h^{(4)}(x) \)

Develop a general rule for \(f^{(n)}(x)\)

From the previous four steps, a pattern emerges. It appears that each derivative can be written as a sum of terms in which the exponents of the derivatives of \(g(x)\) and \(h(x)\) add up to the order of the derivative. The coefficient of each term appears to be a binomial coefficient, that is, the coefficients from the binomial theorem that appear in the expansion of \((a + b)^n\). Thus, we can conjecture a general rule for \(f^{(n)}(x)\), which should be carefully verified: \(f^{(n)}(x) = ∑_{k=0}^{n} C(n, k) g^{(n-k)}(x) h^{(k)}(x)\), where \(C(n, k)\) is the binomial coefficient 'n choose k', which denotes the number of ways to choose \(k\) elements from a set of \(n\) elements.