How do you find the equation for the tangent line for #y=1/x# at #x=1#?

Question

Cameron Alexander · Accepted Answer

#y=1/x# can be expressed in an alternate form as
#y=x^(-1)# The derivative of #y# gives the slope of the tangent line at a point #(x,y)# #(dy)/(dx) = (-1)x^(-2)#
or (equivalently)
#(dy)/(dx) = - 1/x^2# At x = 1
the slope is #(-1)/(1^2) = -1# at #x=1#
#y = 1/x = 1# The tangent has a slope of #(-1)# and passes through the point #(1,1)# Using the slope-point form
#(y-1) = (-1)(x-1)#
#y-1 = 1-x#
#y=2-x# The equation for the required tangent line is
#y=2-x#

Cameron Alexander · Answer

You can alternatively use the Newton Approximation Method, which is: #y = f(a) + f'(a)(x-a)#
where a is some arbitrary value and x is the typical unknown coordinate. EX: If #f(x) = 1/x#:
#f(1) = 1/1# and #f'(1) = -1/x^2# at #x = 1#
#=> -1/1^2#.  So:
#y = 1/1 -(x-1)/1^2#  or #y = 1 - (x - 1) = 2 - x# is your tangent line. Power Rule: #(f'(1/x) = f'(x^-1) = -1*x^-2 = -1/x^2)#

Scarlett Allison · Answer

undefined To find the equation for the tangent line of the function y = 1/x at x = 1, we need to determine the slope of the tangent line at that point and then use the point-slope form of a linear equation.

First, we find the derivative of the function y = 1/x using the power rule for differentiation. The derivative of 1/x is -1/x^2.

Next, we substitute x = 1 into the derivative to find the slope of the tangent line at x = 1. Plugging in x = 1, we get -1/1^2 = -1.

Now, we have the slope of the tangent line, which is -1. We also have the point (1, 1) since x = 1 is the point of tangency.

Using the point-slope form of a linear equation, y - y1 = m(x - x1), we substitute the values we have:

y - 1 = -1(x - 1)

Simplifying, we get:

y - 1 = -x + 1

Rearranging the equation, we obtain the equation for the tangent line:

y = -x + 2