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

The derivative of (3x^2/(y^2)) is (6x(-x\frac{dy)/(dx)+y)}{y^3}

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

The first derivative of (3x^2/(y^2)) is (6x(-x\frac{dy)/(dx)+y)}{y^3}

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

\frac{6 x ( - x \frac{d y}{d x} + y )}{y ^{3}}

Solution steps

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

Take the constant out:

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

= 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}}

= 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}

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

Simplify

3 \cdot \frac{2 x y ^{2} - 2 y \frac{d y}{d x} x ^{2}}{( y ^{2} ) ^{2}} : \frac{6 x ( - x \frac{d y}{d x} + y )}{y ^{3}}

= \frac{6 x ( - x \frac{d y}{d x} + y )}{y ^{3}}

\int cos (x) sec (x) d x

x \to 0 lim (\frac{1}{1 + 2 ^{\frac{1}{x}}})

\int_{0}^{2} 6 x d x

\int sec (8 x) tan (8 x) d x

(\frac{- x ^{2} - 4 x}{( x ^{2} - x - 2 ) ^{2}})^{^{'}}

What is the derivative of (3x^2/(y^2)) ?
The derivative of (3x^2/(y^2)) is (6x(-x\frac{dy)/(dx)+y)}{y^3}
What is the first derivative of (3x^2/(y^2)) ?
The first derivative of (3x^2/(y^2)) is (6x(-x\frac{dy)/(dx)+y)}{y^3}