What is (\partial)/(\partial x)((2x^2+7x^2y)/(3xy)) ?

The answer to (\partial)/(\partial x)((2x^2+7x^2y)/(3xy)) is (7y+2)/(3y)

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(\partial)/(\partial x)((2x^2+7x^2y)/(3xy))

\frac{\partial}{\partial x} (\frac{2 x ^{2} + 7 x ^{2} y}{3 x y})

\frac{7 y + 2}{3 y}

Solution steps

\frac{\partial}{\partial x} (\frac{2 x ^{2} + 7 x ^{2} y}{3 x y})

Treat

y

as a constant

Take the constant out:

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

= \frac{1}{3 y} \frac{\partial}{\partial x} (\frac{2 x ^{2} + 7 x ^{2} y}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{3 y} \cdot \frac{\frac{\partial}{\partial x} ( 2 x ^{2} + 7 x ^{2} y ) x - \frac{\partial}{\partial x} ( x ) ( 2 x ^{2} + 7 x ^{2} y )}{x ^{2}}

\frac{\partial}{\partial x} (2 x^{2} + 7 x^{2} y) = 4 x + 14 y x

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

= \frac{1}{3 y} \cdot \frac{( 4 x + 14 y x ) x - 1 \cdot ( 2 x ^{2} + 7 x ^{2} y )}{x ^{2}}

Simplify

\frac{1}{3 y} \cdot \frac{( 4 x + 14 y x ) x - 1 \cdot ( 2 x ^{2} + 7 x ^{2} y )}{x ^{2}} : \frac{7 y + 2}{3 y}

= \frac{7 y + 2}{3 y}

x \to 6 lim (\frac{1}{x - 6})

n = 0 \sum \infty 1 8^{n} x^{n} n!

(ln ∣ x ∣)^{^{'}}

f (x) = x^{11} arccot (x)

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

What is (\partial)/(\partial x)((2x^2+7x^2y)/(3xy)) ?
The answer to (\partial)/(\partial x)((2x^2+7x^2y)/(3xy)) is (7y+2)/(3y)