Question

differentiate the function.
f(x)=2x ln(9x)-2x
f(x)=

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differentiate the function.
f(x)=sin(9 ln(x))
f(x)=

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Explanation:

Step1: Differentiate first - term of \(f(x) = 2x\ln(9x)-2x\) using product rule

The product rule \((uv)^\prime=u^\prime v + uv^\prime\), where \(u = 2x\), \(u^\prime=2\), \(v=\ln(9x)\), \(v^\prime=\frac{9}{9x}=\frac{1}{x}\). So \((2x\ln(9x))^\prime=2\ln(9x)+2x\cdot\frac{1}{x}=2\ln(9x) + 2\).

Step2: Differentiate second - term of \(f(x)\)

The derivative of \(-2x\) is \(-2\).

Step3: Find \(f^\prime(x)\)

\(f^\prime(x)=(2x\ln(9x))^\prime-(2x)^\prime=2\ln(9x)+2 - 2=2\ln(9x)\)

Step4: Differentiate \(f(x)=\sin(9\ln(x))\) using chain - rule

Let \(u = 9\ln(x)\), then \(f(x)=\sin(u)\). The chain - rule states that \(f^\prime(x)=\frac{df}{du}\cdot\frac{du}{dx}\). \(\frac{df}{du}=\cos(u)=\cos(9\ln(x))\), \(\frac{du}{dx}=\frac{9}{x}\). So \(f^\prime(x)=\frac{9\cos(9\ln(x))}{x}\)

Answer:

1. \(2\ln(9x)\)
2. \(\frac{9\cos(9\ln(x))}{x}\)