Question

If y = Tan(x)^Tan(x) then ^dy⁄dx⁡ = ?

(a) Tan(x) [1 + lnTan(x)] Tan(x)^Tan(x)

(b) Tan^2 (x) [1 + lnTan(x)] Tan(x)^Tan(x)

(c) Sec^2 (x) [1 + lnTan(x)] Tan(x)^Tan(x)

(d) Sec(x) [1 + lnTan(x)] Tan(x)^Tan(x)

I got this question during an interview.

This key question is from Limits and Derivatives of Several Variables in chapter Partial Differentiation of Engineering Mathematics

Answer 1

Right answer is (c) Sec^2 (x) [1 + lnTan(x)] Tan(x)^Tan(x)

Best explanation: y = Tan(x)^Tan(x)

Taking ln on both side

ln y = Tan(x)lnTan(x)

Differentiating w.r.t x

\(\frac{1}{y} \frac{dy}{dx} = \frac{Tan(x)Sec^2 (x)}{Tan(x)}+Sec^2 (x)lnTan(x)\)

\(\frac{1}{y} \frac{dy}{dx} = Sec^2 (x)[1+lnTan(x)]\)

\(\frac{dy}{dx}=Sec^2 (x)[1+lnTan(x)]Tan(x)^{Tan(x)}\)