#2015_13 I just started learning calculus

Calculus Level 2

( tan x ) ( tan x ) ( tan x ) . . . \large { (\tan { x } ) }^{ { (\tan { x }) }^{ { (\tan { x }) }^{ { . }^{ .{ }^{ . } } } } }

Find the value of derivative of the function above at x = π 4 x=\frac { \pi }{ 4 } .

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The answer is 2.

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1 solution

First note that it is valid to consider this infinite tetration in the neighborhood of x = π 4 , x = \frac{\pi}{4}, and in this neighborhood y = f ( x ) y = f(x) is both continuous and differentiable.

As such, we then observe that, (since y > 0 y \gt 0 in this neighborhood),

ln ( y ) = y ln ( tan ( x ) ) ln ( y ) y = ln ( tan ( x ) ) . \ln(y) = y*\ln(\tan(x)) \Longrightarrow \dfrac{\ln(y)}{y} = \ln(\tan(x)).

Differentiating both sides (using the quotient and chain rules) we have that

1 ln ( y ) y 2 d y d x = sec 2 ( x ) tan ( x ) d y d x = y 2 sec ( x ) csc ( x ) 1 ln ( y ) . \dfrac{1 - \ln(y)}{y^{2}} * \dfrac{dy}{dx} = \dfrac{\sec^{2}(x)}{\tan(x)} \Longrightarrow \dfrac{dy}{dx} = \dfrac{y^{2}\sec(x)\csc(x)}{1 - \ln(y)}.

Now at x = π 4 tan ( x ) = 1 x = \frac{\pi}{4} \Longrightarrow \tan(x) = 1 we have that

y ( 1 ) = 1 y(1) = 1 and sec ( x ) = csc ( x ) = 2 , \sec(x) = \csc(x) =\sqrt{2}, thus

f ( π 4 ) = 1 2 1 0 = 2 . f'(\frac{\pi}{4}) = \dfrac{1*2}{1 - 0} = \boxed{2}.

Yup, my solution was almost the same as yours, sir, but I did not use quotient rule, but in the end, the derivative was same. But, my question is: Is it possible to evaluate the derivative using the limit definition of the derivative?

Abhinandan Padhi - 5 years, 11 months ago

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