What is the d/(dy)(y^y) ?

The d/(dy)(y^y) is y^y(ln(y)+1)

What is the first d/(dy)(y^y) ?

The first d/(dy)(y^y) is y^y(ln(y)+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d y} (y^{y})

y^{y} (ln (y) + 1)

Solution steps

\frac{d}{d y} (y^{y})

Apply exponent rule:

a^{b} = e^{b l n (a)}

y^{y} = e^{y l n (y)}

= \frac{d}{d y} (e^{y l n (y)})

Apply the chain rule:

e^{y l n (y)} \frac{d}{d y} (y ln (y))

= e^{y l n (y)} \frac{d}{d y} (y ln (y))

\frac{d}{d y} (y ln (y)) = ln (y) + 1

= e^{y l n (y)} (ln (y) + 1)

e^{y l n (y)} = y^{y}

= y^{y} (ln (y) + 1)

\frac{d}{d x} (e^{3 x^{2}} + ln (5 x - 4))

\frac{d}{d x} (x - sin^{2} (x) - 4)

\frac{4 sin ( x )}{6 - 7 sin ( x )}

\frac{\partial}{\partial x} (e^{- x} cos (- 8 y))

f (x) = \frac{12}{ln ( x )}