Question

find f(x).
f(x)=6\ln(1 + 9x^{2})
f(x)=\square

Explanation:

Step1: Apply the chain rule

The derivative of \(y = \ln(u)\) is \(y^\prime=\frac{u^\prime}{u}\). Let \(u = 1 + 9x^{2}\), then \(f(x)=6\ln(u)\). The derivative of \(f(x)\) with respect to \(x\) is \(f^\prime(x)=6\times\frac{d}{dx}(\ln(u))\). By the chain rule, \(f^\prime(x)=6\times\frac{u^\prime}{u}\).

Step2: Find \(u^\prime\)

Differentiate \(u = 1+9x^{2}\) with respect to \(x\). Using the power rule \(\frac{d}{dx}(ax^{n})=nax^{n - 1}\), we have \(u^\prime=\frac{d}{dx}(1)+\frac{d}{dx}(9x^{2})=0 + 18x=18x\).

Step3: Substitute \(u\) and \(u^\prime\) into the formula

Substitute \(u = 1 + 9x^{2}\) and \(u^\prime=18x\) into \(f^\prime(x)=6\times\frac{u^\prime}{u}\). Then \(f^\prime(x)=6\times\frac{18x}{1 + 9x^{2}}\).

Step4: Simplify the expression

\(f^\prime(x)=\frac{108x}{1 + 9x^{2}}\)

Answer:

\(\frac{108x}{1 + 9x^{2}}\)