Question

differentiate the function with respect to the independent variable.
f(x)=\ln(8x + 1)
f(x)=\square

Explanation:

Step1: Apply the chain rule

Let \(u = 8x+1\), then \(f(x)=\ln(u)\). The derivative of \(\ln(u)\) with respect to \(u\) is \(\frac{1}{u}\), and the derivative of \(u = 8x + 1\) with respect to \(x\) is \(8\).
By the chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\), we have \(f^\prime(x)=\frac{1}{u}\cdot8\).

Step2: Substitute \(u\) back

Since \(u = 8x+1\), substituting \(u\) into the expression from Step 1 gives \(f^\prime(x)=\frac{8}{8x + 1}\).

Answer:

\(\frac{8}{8x+1}\)