Question

1. find the equation of the line tangent to the graph of ( y=ln \frac{x}{3} ) at ( x = e ).

Explanation:

Step1: Find the derivative of \(y = \ln\frac{x}{3}\)

Using the chain - rule, \(y=\ln x-\ln3\). The derivative of \(\ln x\) is \(\frac{1}{x}\) and the derivative of a constant \(\ln3\) is \(0\). So \(y^\prime=\frac{1}{x}\).

Step2: Evaluate the derivative at \(x = e\)

Substitute \(x = e\) into \(y^\prime\). Then \(y^\prime|_{x = e}=\frac{1}{e}\).

Step3: Find the \(y\) - value when \(x = e\)

Substitute \(x = e\) into \(y=\ln\frac{x}{3}\). So \(y=\ln\frac{e}{3}=1 - \ln3\).

Step4: Use the point - slope form \(y - y_1=m(x - x_1)\)

Here \(m=\frac{1}{e}\), \(x_1 = e\), \(y_1=1-\ln3\).

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Answer:

\(y=\frac{1}{e}x-\ln3\)