What is (\partial)/(\partial x)(2x^2ln(3xy^2+4y)) ?

The answer to (\partial)/(\partial x)(2x^2ln(3xy^2+4y)) is 2(2xln(3xy^2+4y)+(3x^2y)/(3xy+4))

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(\partial)/(\partial x)(2x^2ln(3xy^2+4y))

\frac{\partial}{\partial x} (2 x^{2} ln (3 x y^{2} + 4 y))

2 (2 x ln (3 x y^{2} + 4 y) + \frac{3 x ^{2} y}{3 x y + 4})

Solution steps

\frac{\partial}{\partial x} (2 x^{2} ln (3 x y^{2} + 4 y))

Treat

y

as a constant

Take the constant out:

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

= 2 \frac{\partial}{\partial x} (x^{2} ln (3 x y^{2} + 4 y))

Apply the Product Rule:

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

f = x^{2}, g = ln (3 x y^{2} + 4 y)

= 2 (\frac{\partial}{\partial x} (x^{2}) ln (3 x y^{2} + 4 y) + \frac{\partial}{\partial x} (ln (3 x y^{2} + 4 y)) x^{2})

\frac{\partial}{\partial x} (x^{2}) = 2 x

\frac{\partial}{\partial x} (ln (3 x y^{2} + 4 y)) = \frac{3 y}{3 y x + 4}

= 2 (2 x ln (3 x y^{2} + 4 y) + \frac{3 y}{3 y x + 4} x^{2})

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= 2 (2 x ln (3 x y^{2} + 4 y) + \frac{3 x ^{2} y}{3 x y + 4})

\frac{d}{d x} (x e^{- \frac{x}{2}})

\int_{0}^{5} π (5 x)^{2} d x

\int_{1}^{3} (2 x + 1) d x

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

\int \frac{1}{sec ^{4} ( \frac{1}{4} β )} d β

What is (\partial)/(\partial x)(2x^2ln(3xy^2+4y)) ?
The answer to (\partial)/(\partial x)(2x^2ln(3xy^2+4y)) is 2(2xln(3xy^2+4y)+(3x^2y)/(3xy+4))