What is the derivative of (3x^2/(2y^2)) ?

The derivative of (3x^2/(2y^2)) is (3x(-x\frac{dy)/(dx)+y)}{y^3}

What is the first derivative of (3x^2/(2y^2)) ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{3}{2} \cdot \frac{2 x y ^{2} - 2 y \frac{d y}{d x} x ^{2}}{( y ^{2} ) ^{2}} : \frac{3 x ( - x \frac{d y}{d x} + y )}{y ^{3}}

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

\frac{\partial}{\partial x} (cos (z))

\frac{d}{d x} (e^{0.5} + 5 x - 5)

y^{^{''}} + 10 y + 25 y = 0

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

x y^{^{'}} + y = 3 cos (x), y (π) = 0

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