How do you find the equation of the tangent line to the graph of #y = (ln x)/x# at the points (1,0)?

Question

Ezra Adams · Accepted Answer

The equation is #y = x - 1#.

Differentiate using the quotient rule. #y = lnx/x# #y' = (1/x xx x - lnx xx 1)/x^2# #y' = (1 - lnx)/x^2# Now, we find the slope of the tangent by inserting the point #x = 1# into the derivative. #m_"tangent" = (1- ln(1))/1^2# #m_"tangent" = (1 - 0)/1# #m_"tangent" = 1# We can now find the equation, because we know the slope and a point. #y - y_1 = m(x - x_1)# #y - 0 = 1(x- 1)# #y - 0 = x - 1# #y = x - 1# Hopefully this helps!

William Abdalla · Answer

undefined To find the equation of the tangent line to the graph of y = (ln x)/x at the point (1,0), we need to find the slope of the tangent line at that point.

To find the slope, we can use the derivative of the function y = (ln x)/x.

The derivative of y = (ln x)/x can be found using the quotient rule.

The derivative is given by:

dy/dx = (x * (1/x) - ln x * 1) / x^2

Simplifying this expression, we get:

dy/dx = (1 - ln x) / x^2

Now, substitute x = 1 into the derivative expression to find the slope at the point (1,0):

dy/dx = (1 - ln 1) / 1^2

Since ln 1 = 0, the slope at the point (1,0) is:

dy/dx = (1 - 0) / 1^2

dy/dx = 1

Therefore, the slope of the tangent line at the point (1,0) is 1.

Using the point-slope form of a line, we can write the equation of the tangent line:

y - y1 = m(x - x1)

Substituting the values of the point (1,0) and the slope m = 1, we get:

y - 0 = 1(x - 1)

Simplifying this equation, we have:

y = x - 1

Therefore, the equation of the tangent line to the graph of y = (ln x)/x at the point (1,0) is y = x - 1.