How do I find the tangent line of this function at x=1? y = f(x) = ln(x^2 + 4x + 1) Please show working!! I'm super confused

1 Answer
May 14, 2018

#y=x+ln6-1#

Explanation:

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To find the slope of the tangent to the curve at a specific point, we take the derivative of its function and evaluate it at that point:

#y=ln(x^2+4x+1)#

Let #u=x^2+4x+1, :. (du)/dx=2x+4, and y=lnu#

#(dy)/(du)=(d(lnu))/(du)=1/u#

#dy/dx=(dy)/(du)*(du)/(dx)#

#dy/dx=1/u*(2x+4)=(2x+4)/(x^2+4x+1)#

Now, we evaluate this derivative for #x=1#

#Slope=m=(2(1)+4)/((1)^2+4(1)+1)=6/6=1#

Now, we write the equation of the straight line tangent to the curve, which is in the form of :

#y=mx+b#

#y=(1)x+b#

#y=x+b#

#b# is the #y#-intercept of the line and can be found by plugging in the coordinates of the tangency point. But we need to find the #y#-coordinate of the point of tangency first by plugging #x=1# into the function:

#y=ln6#

Therefore, the point of tangency has the coordinates of:

#(1, ln6)#

Now we plug these into the equation of the tangent line:

#ln6=1+b#

#b=ln6-1#

Therefore, the equation of the tangent line is:

#y=x+ln6-1#

Here are the graphs of the function and the tangent at #x=1#:

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