Question

differentiate

f(x) = x^{5} ln 7x

f(x) =
(use parentheses to clearly denote the argument of each function)

Explanation:

Step1: Apply product rule

The product rule states that if \(y = u\cdot v\), then \(y^\prime=u^\prime v + uv^\prime\). Let \(u = x^{5}\) and \(v=\ln(7x)\). First, find \(u^\prime\): \(u^\prime=\frac{d}{dx}(x^{5}) = 5x^{4}\). Then, find \(v^\prime\) using the chain - rule. Let \(t = 7x\), \(\frac{d}{dx}(\ln(7x))=\frac{d}{dt}(\ln t)\cdot\frac{dt}{dx}\). Since \(\frac{d}{dt}(\ln t)=\frac{1}{t}\) and \(\frac{dt}{dx}=7\), then \(v^\prime=\frac{7}{7x}=\frac{1}{x}\).

Step2: Substitute into product rule formula

\(f^\prime(x)=u^\prime v+uv^\prime\). Substitute \(u = x^{5}\), \(u^\prime = 5x^{4}\), \(v=\ln(7x)\) and \(v^\prime=\frac{1}{x}\) into the formula:

$$
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Answer:

\(x^{4}(5\ln(7x)+ 1)\)