What is (\partial)/(\partial y)(x^y-xy) ?

The answer to (\partial)/(\partial y)(x^y-xy) is x^yln(x)-x

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(\partial)/(\partial y)(x^y-xy)

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

x^{y} ln (x) - x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

\frac{d y}{d x} = 2 x e^{y}

\frac{d}{d x} ((3 x)^{3 x})

\int \frac{1}{a + b x} d x

\frac{d}{d x} (- 4 cot (2 x) csc (2 x))

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 sin (2 t)

What is (\partial)/(\partial y)(x^y-xy) ?
The answer to (\partial)/(\partial y)(x^y-xy) is x^yln(x)-x