How do you find the equation of the tangent line to #f(x)=1/x# at the point (1,1)?

1 Answer
May 15, 2015

The "how" depends partly on how much calculus you have already learned.
This answer is long because I've talked about finding the slope 3 different ways,

Typically, we have students do this kind of problem first using the definition of slope of a tangent line at #a#.
Then we give the more general definition of the derivative of a function and have students use that to find the slope of the tangent.
Later still, we develop the general rules (properties) for derivatives that simplify the problem considerably.

If you are at the first stage, use:

slope of the tangent at #x=1# is

#m=lim_(xrarr1)(f(x)-f(1))/(x-1)#, so

#m=lim_(xrarr1) (1/x-1/1)/((x-1)/1) #

#color(white)"ss"# #=lim_(xrarr1) ((1-x)/x*1/(x-1)) #

#color(white)"ss"# #=lim_(xrarr1) ((-(x-1))/x*1/(x-1)) #

#color(white)"ss"# #=lim_(xrarr1) 1/x =1#

So #m=1#.

The definition of derivative is similar (Here is Gio's answer to that question for this function. )

If you have learned the power rule for negative exponents, use:

#f(x)=1/x=x^(-1)#, so the derivative is #f'(x)=(-1)x^(-1-1) = -1/x^2#

And the slope of the tangent at #1# is #f'(1)=-1/(1)^2 = -1#.

Finally, write the equation of the line through #(1,1)# with slope #-1#

#y-1=-1(x-1)# so

#y=-x+2#