Using the Tangent Line to Approximate Function Values
Key Questions

A formula for a tangent line approximation of a function f, also called linear approximation , is given by
#f(x)~~f(a)+f'(a)(xa),#
which is a good approximation for#x# when it is close enough to#a# .I'm not sure, but I think the question is about approximate the value
#ln(1.004)# . Could you verify it please? Otherwise we will need to know an approximation to#ln(10).# In this case, we have
#f(x)=ln(x)# ,#x=1.004# and#a=1# .Since,
#f'(x)=1/x=>f'(1)=1# and#ln(1)=0# , we get#ln(1.004)~~ln(1)+1*(1.0041)=0.004.# 
We need to find the derivative of
#f(x)# . We need to use the Chain Rule to find the derivative of#f(x)# .#f(x)=sqrt(1+x)=(1+x)^(1/2)# #f'(x)=(1/2)(1+x)^((1/21))*1# #f'(x)=(1/2)(1+x)^((1/22/2))# #f'(x)=(1/2)(1+x)^((1/2))# #f'(x)=1/(2sqrt(1+x))# #f'(0)=1/(2sqrt(1+0))=1/(2sqrt(1))=1/2=0.5# 
The linear approximation
#L(x)# of a function is done using the tangent line to the graph of the function. The equation of the tangent line at#x=a# is given by#y=f'(a)(xa)+f(a)# ,the linear approximation is
#L(x)=f'(a)(xa)+f(a)# .