How do you find the linearization at a=1 of #f(x) = 2 ln(x)#? Calculus Applications of Derivatives Using the Tangent Line to Approximate Function Values 1 Answer Wataru Apr 1, 2017 #L(x)=2(x-1)# Explanation: #f(x)=2ln x Rightarrow f(1)=2ln(1)=0# # f'(x)=2/x Rightarrow f'(1)=2/1=2# Linearization #L(x)# of #f# at #a=1# is: #L(x)=f(a)+f'(a)(x-a)=f(1)+f'(1)(x-1)=2(x-1)# I hope that this was clear. Answer link Related questions How do you find the linear approximation of #(1.999)^4# ? How do you find the linear approximation of a function? How do you find the linear approximation of #f(x)=ln(x)# at #x=1# ? How do you find the tangent line approximation for #f(x)=sqrt(1+x)# near #x=0# ? How do you find the tangent line approximation to #f(x)=1/x# near #x=1# ? How do you find the tangent line approximation to #f(x)=cos(x)# at #x=pi/4# ? How do you find the tangent line approximation to #f(x)=e^x# near #x=0# ? How do you use the tangent line approximation to approximate the value of #ln(1003)# ? How do you use the tangent line approximation to approximate the value of #ln(1.006)# ? How do you use the tangent line approximation to approximate the value of #ln(1004)# ? See all questions in Using the Tangent Line to Approximate Function Values Impact of this question 2160 views around the world You can reuse this answer Creative Commons License