What is (\partial)/(\partial y)(-e^xln(y)) ?

The answer to (\partial)/(\partial y)(-e^xln(y)) is -(e^x)/y

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(\partial)/(\partial y)(-e^xln(y))

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

- \frac{e ^{x}}{y}

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the common derivative:

\frac{\partial}{\partial y} (ln (y)) = \frac{1}{y}

= - e^{x} \frac{1}{y}

Simplify

- e^{x} \frac{1}{y} : - \frac{e ^{x}}{y}

= - \frac{e ^{x}}{y}

y = sin (sin (x)), (π, 0)

\frac{d}{d x} (\frac{8 x ^{2}}{2 x - 10})

\frac{d}{d x} (\frac{x}{t})

5 \to 1 lim (\frac{\frac{1}{5}}{5 - 1})

y = \frac{8}{x}, y = 2 x, x = 4

What is (\partial)/(\partial y)(-e^xln(y)) ?
The answer to (\partial)/(\partial y)(-e^xln(y)) is -(e^x)/y