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}}

\int - 9 sin (x) d x

f (x) = x + x

\int \frac{1}{x - 2 x ^{\frac{3}{4}} - 8 x} d x

n = 0 \sum \infty \frac{5 + n}{1 0 ^{n}}

\int (\frac{csc ( x ) + cot ( x )}{sin ( x )}) d 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}