What is the derivative of (3y^2/(x^2-4)) ?

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

What is the first derivative of (3y^2/(x^2-4)) ?

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

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

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

Solution steps

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

Take the constant out:

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

= 3 \frac{d}{d x} (\frac{y ^{2}}{x ^{2} - 4})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{2} - 4) = 2 x

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 ( 2 y \frac{d y}{d x} ( x ^{2} - 4 ) - 2 x y ^{2} )}{( x ^{2} - 4 ) ^{2}}

f (x) = 3 x^{4} - 8 x^{3} + 6 x^{2}

f (x) = 2 x^{3} - 2 x^{2} - 16 x + 1

\frac{2 x - 1}{x - 1}

f (x) = x^{2} - 6 x + 7

f (x) = 2 x^{3} + 6 x^{2} - 18 x - 4

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