How do you write the equation of the tangent line to the graph of #y = ln(x^2)# at the point (5, ln(25))?
1 Answer
Mar 29, 2016
One fast way to do it is using the linear approximation method. For this, we have:
#color(blue)(f_T(x) = f(a) + f'(a)(x-a))# where:
#f_T(x)# is the function for the tangent line.#f(a)# is the function#f(x)# at#x = a# .#f'(a)# is the derivative,#f'(x)# , at#x = a# .#a# is the value of#\mathbf(x)# that you are using.
Since
#color(green)(f'(x)) = 1/(x^2)*2x#
#= color(green)(2/x)# (remember to use the Chain Rule on#x^2# )
Next, we can find
- Since we are using
#x = a# , we have#color(green)(a = 5)# from#(5,ln25)# . #f(a)# is#f(5)# . so#color(green)(f(5) = ln25)# , as given in the coordinate#(5,ln25)# .- For
#f'(a)# , we plug in#x = a = 5# to get#color(green)(f'(5)) = 2/a = color(green)(2/5)# .
So, the tangent line becomes:
#color(blue)(f_T(x)) = ln25 + 2/5(x-5)#
#= ln25 + 2/5x - 2#
#= stackrel("m")overbrace(color(blue)(2/5))stackrel("x")overbrace(color(blue)(x)) color(blue)(+) stackrel("b")overbrace(color(blue)(ln25 - 2))#
