What is (\partial)/(\partial x)(amx^bmy^{1-b}) ?

The answer to (\partial)/(\partial x)(amx^bmy^{1-b}) is am^2y^{1-b}bx^{b-1}

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(\partial)/(\partial x)(amx^bmy^{1-b})

\frac{\partial}{\partial x} (am x^{b} m y^{1 - b})

a m^{2} y^{1 - b} b x^{b - 1}

Solution steps

\frac{\partial}{\partial x} (am x^{b} m y^{1 - b})

Simplify

am x^{b} m y^{1 - b} : a m^{2} x^{b} y^{1 - b}

= \frac{\partial}{\partial x} (a m^{2} x^{b} y^{1 - b})

Treat

a, m, b, y

as constants

Take the constant out:

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

= a m^{2} y^{1 - b} \frac{\partial}{\partial x} (x^{b})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= a m^{2} y^{1 - b} b x^{b - 1}

\frac{d}{d x} (arccos (x^{2}))

\int_{0}^{1} \frac{2 x}{e ^{2 x}} d x

y = 2 x^{3} + 3 x^{2} - 12 x + 7

\frac{d}{d x} (e^{2 x} (2 x + 1))

h \to 0 lim (\frac{49 + h - 7}{h})

What is (\partial)/(\partial x)(amx^bmy^{1-b}) ?
The answer to (\partial)/(\partial x)(amx^bmy^{1-b}) is am^2y^{1-b}bx^{b-1}