Question

if y=x^x find dy/dx​

Answer 1

Step-by-step explanation:

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)

Hope it helps!

Answer 2

Solution:

$y = x {}^{x}$

take log on both sides

$log(y) = x log(x)$

now defrentitate with respect to x

by chain rule

$y \frac{dy}{dx} = x \times \frac{1}{x} + log(x)$

$y \frac{dy}{dx} = 1 + log(x)$

$\frac{dy}{dx} = y(1 + log(x) )$

$\frac{dy}{dx} = x {}^{x} (1 + log(x) )$

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