Math CentralQuandaries & Queries


Question from alex, a student:

y = log2(log2x)

The slope of the tangent to the given curve at its x-intercept is..?


I would start by reminding myself of the definition of lag base 2.

s = log2(t) if and only if t = 2s         (1)

The x-intercept is the point on the curve where y = 0. From (1) is 0 = log2(log2x) then log2(x) = 1 and then from (1) again x = 2. Thus the x-intercept is (2,0).

To find the slope of the tangent to the curve at this point you need to find the derivative of y = log2(log2x) with respect to x. This in going to involve the chain rule and the derivative of log2x. Again using (1),

if s = log2t then t = 2s so ln(t) = ln(2s) = s ln(2)

Differentiate both sides with respect to t to obtain

1/t = ds/dt ln(2)

and hence

if s = log2t then ds/dt = 1/(t ln(2))

Use this fact and the chain rule to differentiate y = log2(log2x) and then evaluate the derivative at (2, 0).


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