What is (\partial)/(\partial x)(y^{x^y}) ?

The answer to (\partial)/(\partial x)(y^{x^y}) is y^{x^y+1}x^{y-1}ln(y)

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(\partial)/(\partial x)(y^{x^y})

\frac{\partial}{\partial x} (y^{x^{y}})

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

Solution steps

\frac{\partial}{\partial x} (y^{x^{y}})

Treat

y

as a constant

Apply exponent rule:

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

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (x^{y} ln (y)) = ln (y) y x^{y - 1}

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

Simplify

e^{x^{y} l n (y)} ln (y) y x^{y - 1} : y^{x^{y} + 1} x^{y - 1} ln (y)

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

\frac{d}{d x} (e^{3 x} cos (4 x))

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{- 4} e^{2 t}

t^{2} \cdot e^{- 3 t}

\frac{d}{d x} (\frac{x}{10} - \frac{10}{x})

\int 20 θ sec^{2} (θ) d θ

What is (\partial)/(\partial x)(y^{x^y}) ?
The answer to (\partial)/(\partial x)(y^{x^y}) is y^{x^y+1}x^{y-1}ln(y)