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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Simplify

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

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

\frac{d}{d x} (1 - 3 x^{2})

x \to 0 + lim (tan (x) \cdot ln (x))

\int - x^{- \frac{1}{2}} d x

\frac{d}{d x} (8 x sin (x) - 4 π sin (x))

\frac{d}{d x} (ln^{- 1} (x))

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