What is (\partial)/(\partial x)(y^{4/x}) ?

The answer to (\partial)/(\partial x)(y^{4/x}) is -(4y^{4/x}ln(y))/(x^2)

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(\partial)/(\partial x)(y^{4/x})

\frac{\partial}{\partial x} (y^{\frac{4}{x}})

- \frac{4 y ^{\frac{4}{x}} ln ( y )}{x ^{2}}

Solution steps

\frac{\partial}{\partial x} (y^{\frac{4}{x}})

Treat

y

as a constant

Apply exponent rule:

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

y^{\frac{4}{x}} = e^{\frac{4}{x} l n (y)}

= \frac{\partial}{\partial x} (e^{\frac{4}{x} l n (y)})

Apply the chain rule:

e^{\frac{4}{x} l n (y)} \frac{\partial}{\partial x} (\frac{4}{x} ln (y))

= e^{\frac{4}{x} l n (y)} \frac{\partial}{\partial x} (\frac{4}{x} ln (y))

\frac{\partial}{\partial x} (\frac{4}{x} ln (y)) = - \frac{4 ln ( y )}{x ^{2}}

= e^{\frac{4}{x} l n (y)} (- \frac{4 ln ( y )}{x ^{2}})

Simplify

e^{\frac{4}{x} l n (y)} (- \frac{4 ln ( y )}{x ^{2}}) : - \frac{4 y ^{\frac{4}{x}} ln ( y )}{x ^{2}}

= - \frac{4 y ^{\frac{4}{x}} ln ( y )}{x ^{2}}

0 \to 0 lim (ln (0))

x \to - \infty lim (\frac{4 - 7 x}{2 + 3 x})

\frac{\partial}{\partial y} (x ln (x) + y ln (x))

\int \frac{1}{b - a y} d y

\int (x^{2} - 4 x + 5) d x

What is (\partial)/(\partial x)(y^{4/x}) ?
The answer to (\partial)/(\partial x)(y^{4/x}) is -(4y^{4/x}ln(y))/(x^2)