What is (\partial)/(\partial x)(xy(1-7x-4y)) ?

The answer to (\partial)/(\partial x)(xy(1-7x-4y)) is y(-14x-4y+1)

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(\partial)/(\partial x)(xy(1-7x-4y))

\frac{\partial}{\partial x} (x y (1 - 7 x - 4 y))

y (- 14 x - 4 y + 1)

Solution steps

\frac{\partial}{\partial x} (x y (1 - 7 x - 4 y))

Treat

y

as a constant

Take the constant out:

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

= y \frac{\partial}{\partial x} (x (1 - 7 x - 4 y))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = 1 - 7 x - 4 y

= y (\frac{\partial}{\partial x} (x) (1 - 7 x - 4 y) + \frac{\partial}{\partial x} (1 - 7 x - 4 y) x)

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (1 - 7 x - 4 y) = - 7

= y (1 \cdot (1 - 7 x - 4 y) + (- 7) x)

Simplify

y (1 \cdot (1 - 7 x - 4 y) + (- 7) x) : y (- 14 x - 4 y + 1)

= y (- 14 x - 4 y + 1)

8 cos (2 x), 8 - 8 cos (2 x), 0, \frac{π}{2}

x \to 0 + lim (x - \frac{1}{x ^{3}})

\frac{\partial}{\partial x} (ln (1 + x^{2} + e^{y} + z))

y = sec (x), (\frac{π}{6}, \frac{2 3}{3})

2 x y^{^{'}} - 8 y = x^{- 4}

What is (\partial)/(\partial x)(xy(1-7x-4y)) ?
The answer to (\partial)/(\partial x)(xy(1-7x-4y)) is y(-14x-4y+1)