What is (\partial)/(\partial x)((x-a/(y^3))(b-x)) ?

The answer to (\partial)/(\partial x)((x-a/(y^3))(b-x)) is b-2x+a/(y^3)

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(\partial)/(\partial x)((x-a/(y^3))(b-x))

\frac{\partial}{\partial x} ((x - \frac{a}{y ^{3}}) (b - x))

b - 2 x + \frac{a}{y ^{3}}

Solution steps

\frac{\partial}{\partial x} ((x - \frac{a}{y ^{3}}) (b - x))

Treat

a, y, b

as constants

Apply the Product Rule:

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

f = x - \frac{a}{y ^{3}}, g = b - x

= \frac{\partial}{\partial x} (x - \frac{a}{y ^{3}}) (b - x) + \frac{\partial}{\partial x} (b - x) (x - \frac{a}{y ^{3}})

\frac{\partial}{\partial x} (x - \frac{a}{y ^{3}}) = 1

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

= 1 \cdot (b - x) + (- 1) (x - \frac{a}{y ^{3}})

Simplify

1 \cdot (b - x) + (- 1) (x - \frac{a}{y ^{3}}) : b - 2 x + \frac{a}{y ^{3}}

= b - 2 x + \frac{a}{y ^{3}}

\frac{d P}{d t} = 7 Pt, P (1) = 5

n = 1 \sum \infty \frac{1}{n} (- 1)^{n}

- y^{^{''}} + 3 y^{^{'}} = 0

y = 2 x^{2} - 6 x + 6, (1, 2)

\int a^{4 x} d x

What is (\partial)/(\partial x)((x-a/(y^3))(b-x)) ?
The answer to (\partial)/(\partial x)((x-a/(y^3))(b-x)) is b-2x+a/(y^3)