What is the derivative of-(2x^2/(5y^2)) ?

The derivative of-(2x^2/(5y^2)) is -(4x(-x\frac{dy)/(dx)+y)}{5y^3}

What is the first derivative of-(2x^2/(5y^2)) ?

The first derivative of-(2x^2/(5y^2)) is -(4x(-x\frac{dy)/(dx)+y)}{5y^3}

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\frac{d}{d x} (- \frac{2 x ^{2}}{5 y ^{2}})

- \frac{4 x ( - x \frac{d y}{d x} + y )}{5 y ^{3}}

Solution steps

\frac{d}{d x} (- \frac{2 x ^{2}}{5 y ^{2}})

Take the constant out:

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

= - \frac{2}{5} \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}{5} \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}{5} \cdot \frac{2 x y ^{2} - 2 y \frac{d y}{d x} x ^{2}}{( y ^{2} ) ^{2}}

Simplify

- \frac{2}{5} \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 )}{5 y ^{3}}

= - \frac{4 x ( - x \frac{d y}{d x} + y )}{5 y ^{3}}

y^{^{'}} = \frac{( 1 - x )}{y}

(u^{- 2} + u^{- 3}) (u^{5} + 5 u^{2})

\frac{d}{d x} (tan^{2} (x θ))

\int_{1}^{e^{5}} \frac{5}{x} d x

f (x) = 1 + 4 - x, at x = 2

What is the derivative of-(2x^2/(5y^2)) ?
The derivative of-(2x^2/(5y^2)) is -(4x(-x\frac{dy)/(dx)+y)}{5y^3}
What is the first derivative of-(2x^2/(5y^2)) ?
The first derivative of-(2x^2/(5y^2)) is -(4x(-x\frac{dy)/(dx)+y)}{5y^3}