Discussion

If \(y = (\tan x)^{x}\), then \(\frac{1}{y}\frac{dy}{dx}\) =

A.\(\log(\tan x) + 2x\, cosec(2x)\)
B.\(\log(\tan x) + x\, cosec(2x)\)
C.\(x\log(\tan x) + 2x\, cosec(2x)\)
D.\(x\log(\tan x) + x^{2} cosec(2x)\)
E.\(\log(\tan x) + \frac{x}{2} cosec(2x)\)

Answer: A

\(\frac{1}{y}\frac{dy}{dx} = \log(\tan x) + 2x\, cosec(2x)\).

Take logarithms and differentiate:

\[ \begin{aligned} \log y &= x\log(\tan x) \\ \frac{1}{y}\frac{dy}{dx} &= \log(\tan x) + x\frac{\sec^{2}x}{\tan x} \end{aligned} \]

and \(\frac{\sec^{2}x}{\tan x} = \frac{1}{\sin x\cos x}\) \(= \frac{2}{\sin 2x} = 2\,cosec\,2x\). So the second term is \(2x\,cosec\,2x\).

Options B and E lose the factor 2 or halve it, and options C and D multiply \(\log(\tan x)\) by an extra x, which the product rule does not produce.

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