How do you find the slope of the line tangent to the graph of #y = ln(x^2)# at the point (4, ln(16))?

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Answer 1 · 2016-08-19T15:08:18

The slope of the tangent is #1/2#

First, differentiate using the chain rule:

Let #y = ln(u)# and #u = x^2#. If so, #dy/dx = 1/u xx 2x = (2x)/(x^2)#.

The slope of the tangent is given by substituting your point, #x = a#, into the derivative.

So, #m_"tangent" = (2(4))/(4^2) = 8/16 = 1/2#

Hence, the slope of the tangent is #1/2#.

Practice exercises:

Determine the slopes of the tangents at the indicated points.

a) #y = log_2(4x^3)# at the point #x = 2#.

b) #y = sqrt(3x^2 - 2x - 1)# at the point #x = 5#

c) #y = 1/(2x^3 + x - 2)# at the point #x = 1#

Hopefully this helps, and good luck!