find equations of the tangent lines to the curve ( y=\frac{(5 ln (x))}{x} ) at the points ( (1,0) ) and (…

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})