What is the derivative of x^{3/4}y^3 ?

The derivative of x^{3/4}y^3 is (3y^3+12xy^2\frac{dy)/(dx)}{4x^{1/4}}

What is the first derivative of x^{3/4}y^3 ?

The first derivative of x^{3/4}y^3 is (3y^3+12xy^2\frac{dy)/(dx)}{4x^{1/4}}

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\frac{d}{d x} (x^{\frac{3}{4}} y^{3})

\frac{3 y ^{3} + 12 x y ^{2} \frac{d y}{d x}}{4 x ^{\frac{1}{4}}}

Solution steps

\frac{d}{d x} (x^{\frac{3}{4}} y^{3})

Apply the Product Rule:

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

f = x^{\frac{3}{4}}, g = y^{3}

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

\frac{d}{d x} (x^{\frac{3}{4}}) = \frac{3}{4 x ^{\frac{1}{4}}}

\frac{d}{d x} (y^{3}) = 3 y^{2} \frac{d y}{d x}

= \frac{3}{4 x ^{\frac{1}{4}}} y^{3} + 3 y^{2} \frac{d y}{d x} x^{\frac{3}{4}}

Simplify

\frac{3}{4 x ^{\frac{1}{4}}} y^{3} + 3 y^{2} \frac{d y}{d x} x^{\frac{3}{4}} : \frac{3 y ^{3} + 12 x y ^{2} \frac{d y}{d x}}{4 x ^{\frac{1}{4}}}

= \frac{3 y ^{3} + 12 x y ^{2} \frac{d y}{d x}}{4 x ^{\frac{1}{4}}}

\int_{0}^{3} ∣ 3 x - 1 ∣ d x

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

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\int x (x^{2} - 3)^{3} d x

What is the derivative of x^{3/4}y^3 ?
The derivative of x^{3/4}y^3 is (3y^3+12xy^2\frac{dy)/(dx)}{4x^{1/4}}
What is the first derivative of x^{3/4}y^3 ?
The first derivative of x^{3/4}y^3 is (3y^3+12xy^2\frac{dy)/(dx)}{4x^{1/4}}