How do you find the equation of a line tangent to the function #y=ln(-x)# at (-2,ln2)?

Question

Emily Ammerman · Accepted Answer

#x+2y+2-2ln2=0# First let us confirm whether #(-2,ln2)# lies on #y=ln(-x)#, Put #x=-2# in #y=ln(-x)#, we get #ln2# and hence it lies on the currve. Now slope of a curve given by #y=f(x)# is given by #(dy)/(dx)# As #y=ln(-x)#, #(dy)/(dx)=1/(-x)xx-1=1/x# and at #x=-2#, #(dy)/(dx)=-1/2# As tangent has a slope #-1/2# and passes through #(-2,ln2)#, its equation is #y-ln2=-1/2(x-(-2))# or #2y-2ln2=-x-2# or #x+2y+2-2ln2=0#

Adam Adkins · Answer

undefined To find the equation of a line tangent to the function y=ln(-x) at (-2,ln2), we need to find the slope of the tangent line at that point.

First, we find the derivative of the function y=ln(-x) using the chain rule. The derivative of ln(-x) is -1/x.

Next, we substitute x=-2 into the derivative to find the slope at that point. The slope is -1/(-2) = 1/2.

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we substitute the values (-2, ln2) and 1/2 into the equation.

Therefore, the equation of the line tangent to the function y=ln(-x) at (-2,ln2) is y - ln2 = (1/2)(x + 2).