Question

find equations of the tangent lines to the curve ( y=\frac{(5 ln (x))}{x} ) at the points ( (1,0) ) and ( left(e, \frac{5}{e}
ight) ) at the point ( (1,0) quad y= ) at the point ( left(e, \frac{5}{e}
ight) quad y= ) illustrate by graphing the curve and its tangent lines.

Explanation:

Step1: Differentiate the function

Given \(y = \frac{5\ln(x)}{x}\), use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u = 5\ln(x)\), \(u^\prime=\frac{5}{x}\), and \(v = x\), \(v^\prime = 1\). Then \(y^\prime=\frac{\frac{5}{x}\cdot x-5\ln(x)\cdot1}{x^{2}}=\frac{5 - 5\ln(x)}{x^{2}}\).

Step2: Find the slope at \((1,0)\)

Substitute \(x = 1\) into \(y^\prime\). \(y^\prime|_{x = 1}=\frac{5-5\ln(1)}{1^{2}}\). Since \(\ln(1)=0\), the slope \(m_1 = 5\).
Using the point - slope form \(y - y_1=m(x - x_1)\), with \((x_1,y_1)=(1,0)\) and \(m = 5\), we get \(y-0 = 5(x - 1)\), so \(y=5x - 5\).

Step3: Find the slope at \((e,\frac{5}{e})\)

Substitute \(x = e\) into \(y^\prime\). \(y^\prime|_{x = e}=\frac{5-5\ln(e)}{e^{2}}\). Since \(\ln(e)=1\), the slope \(m_2=\frac{5 - 5\times1}{e^{2}}=0\).
Using the point - slope form \(y - y_1=m(x - x_1)\), with \((x_1,y_1)=(e,\frac{5}{e})\) and \(m = 0\), we get \(y-\frac{5}{e}=0\times(x - e)\), so \(y=\frac{5}{e}\).

Answer:

At the point \((1,0)\): \(y = 5x-5\)
At the point \((e,\frac{5}{e})\): \(y=\frac{5}{e}\)