What is (\partial)/(\partial y)(x^{1/y}) ?

The answer to (\partial)/(\partial y)(x^{1/y}) is -(x^{1/y}ln(x))/(y^2)

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(\partial)/(\partial y)(x^{1/y})

\frac{\partial}{\partial y} (x^{\frac{1}{y}})

- \frac{x ^{\frac{1}{y}} ln ( x )}{y ^{2}}

Solution steps

\frac{\partial}{\partial y} (x^{\frac{1}{y}})

Treat

x

as a constant

Apply exponent rule:

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

x^{\frac{1}{y}} = e^{\frac{1}{y} l n (x)}

= \frac{\partial}{\partial y} (e^{\frac{1}{y} l n (x)})

Apply the chain rule:

e^{\frac{1}{y} l n (x)} \frac{\partial}{\partial y} (\frac{1}{y} ln (x))

= e^{\frac{1}{y} l n (x)} \frac{\partial}{\partial y} (\frac{1}{y} ln (x))

\frac{\partial}{\partial y} (\frac{1}{y} ln (x)) = - \frac{ln ( x )}{y ^{2}}

= e^{\frac{1}{y} l n (x)} (- \frac{ln ( x )}{y ^{2}})

Simplify

e^{\frac{1}{y} l n (x)} (- \frac{ln ( x )}{y ^{2}}) : - \frac{x ^{\frac{1}{y}} ln ( x )}{y ^{2}}

= - \frac{x ^{\frac{1}{y}} ln ( x )}{y ^{2}}

\frac{d ^{2}}{d x ^{2}} y + 64 y = 0

\frac{\partial}{\partial x} (\frac{2 y}{x ^{2} - y ^{2}})

y^{^{''}} + 10 y^{^{'}} + 25 y = 14 e^{- 5 x}

x \to \infty lim (- \frac{1}{x})

x \to 1 lim (cos (\frac{π}{3}))

What is (\partial)/(\partial y)(x^{1/y}) ?
The answer to (\partial)/(\partial y)(x^{1/y}) is -(x^{1/y}ln(x))/(y^2)