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

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Answer 1 · 2017-11-21T04:47:05

#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#