Question

find dy/dx if y=x^-x​

Answer 1

Given Expression,

$\sf \: y = {x}^{ - x}$

Taking log on both sides,

$\longrightarrow \sf \: log(y) = log( {x}^{ - x} ) \\ \\ \longrightarrow \sf \: log(y) = - x log( x )$

Differentiating w.r.t x,

$\longrightarrow \sf \: \dfrac{ d \{log(y) \} }{dy} \times \dfrac{dy}{dx} = log(x) \dfrac{d( - x)}{dx} - x \dfrac{d \{ log(x) \}}{dx} \\ \\ \longrightarrow \sf \: \dfrac{1}{y} \times \dfrac{dy}{dx} = - log(x) - \dfrac{x}{x} \\ \\ \longrightarrow \sf \: \dfrac{dy}{dx} = - y \bigg \{ log(x) + 1 \bigg \} \\ \\ \longrightarrow \boxed{ \boxed{\sf \: \dfrac{dy}{dx} = - {x}^{ - x} \bigg \{ log(x) + 1 \bigg \} }}$

Answer 2

Answer:

Given : y = xˣ

Apply log on both sides, we get  

log y = log(xˣ)  

⇒ log y = x log x  

Differentiate with respect to x on both sides, we get  

⇒ (d/dx)(log y) = (d/dx)(x log x)  

⇒ (1/y) * (dy/dx) = x(log x)' + (log x)x'  

⇒ (1/y) * (dy/dx) = (x/x) + 1 * log x  

⇒ (1/y) * (dy/dx) = 1 + log x  

⇒ (dy/dx) = y(1 + log x)  

⇒ (dy/dx) = xˣ(1 + log x)  

Step-by-step explanation:

thanks.