find the equation of the tangent line to the graph of y= f(x) = ln (x)/x at the point (1,0)
f'(x) = (1-lnx)/x^2
f'(1) = 1
So, now you have a point and a slope. The line is
y-0 = 1(x-1)
y=x-1
To find the equation of the tangent line to the graph of y = ln(x)/x at the point (1,0), we need to calculate the slope of the tangent line at that point.
The slope of a tangent line to a curve at a given point can be determined using the derivative of the function. So, we need to find the derivative of f(x) = ln(x)/x.
To find the derivative, we can apply the quotient rule or rewrite the function in a different form. Let's use the quotient rule for this example.
The quotient rule states that if we have a function f(x) = g(x) / h(x), then the derivative of f(x), denoted as f'(x), is given by:
f'(x) = [g'(x) * h(x) - g(x) * h'(x)] / [h(x)]^2
Applying the quotient rule to the function f(x) = ln(x)/x, we have:
g(x) = ln(x), g'(x) = 1/x
h(x) = x, h'(x) = 1
f'(x) = [(1/x) * x - ln(x) * 1] / [x^2]
= (1 - ln(x)) / x^2
Now that we have the derivative, we can calculate the slope of the tangent line at x = 1 by substituting x = 1 into f'(x):
f'(1) = (1 - ln(1)) / 1^2
= 1 - 0
= 1
So, the slope of the tangent line at (1,0) is 1.
Next, we use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line, and m is the slope of the line.
Substituting the values of the point (1,0) and the slope m = 1 into the point-slope form, we get:
y - 0 = 1(x - 1)
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.