Question

Do the derivative of:

$x^-x$​

Answer 1

Answer:

$let \: y \: = {x}^{ - x} \\ taking \: log \: both \: sides, \: we \: get, \\ log(y) = log( {x}^{ - x} ) \\ = > log(y) = - x log(x) \\ now \: differentiating \: both \: sides \: w.r.to \: x ,\\ \frac{d}{dx} log(y) = \frac{d}{dx} ( - x log(x)) \\ = > \frac{1}{y} \frac{dy}{dx} = log(x) \times \frac{d}{dx} ( - x) + ( - x) \times \frac{d}{dx} log(x) \\ = > \frac{1}{y} \frac{dy}{dx} = log(x) \times ( - 1) + ( - x) \times \frac{1}{x} \\ = > \frac{1}{y} \frac{dy}{dx} = - log(x) - 1 \\ = > \frac{1}{y} \frac{dy}{dx} = - (1 + logx) \\ = > \frac{dy}{dx} = -y (1 + logx) \\ putting \: the \: value \: of \: y \: in \: the \: above \: eqn: \\ = > \frac{dy}{dx} = - {x}^{ - x} (1 + logx)$

hence, this is the req derivative.

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Formula used :-

$☆ log( {x}^{y} ) = y log(x) \\ ☆ \frac{d}{dx} log(x) = \frac{1}{x} \\ \: \: \: \: \: \:☆ \frac{d}{dx} (uv) = v \times \frac{du}{dx} + u \times \frac{dv}{dx}$