What is the derivative of (6x^2+4/(3y^2-4)) ?

The derivative of (6x^2+4/(3y^2-4)) is (12x(3y^2-4)-6y\frac{dy)/(dx)(6x^2+4)}{(3y^2-4)^2}

What is the first derivative of (6x^2+4/(3y^2-4)) ?

The first derivative of (6x^2+4/(3y^2-4)) is (12x(3y^2-4)-6y\frac{dy)/(dx)(6x^2+4)}{(3y^2-4)^2}

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\frac{d}{d x} (\frac{6 x ^{2} + 4}{3 y ^{2} - 4})

\frac{12 x ( 3 y ^{2} - 4 ) - 6 y \frac{d y}{d x} ( 6 x ^{2} + 4 )}{( 3 y ^{2} - 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x ^{2} + 4}{3 y ^{2} - 4})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 6 x ^{2} + 4 ) ( 3 y ^{2} - 4 ) - \frac{d}{d x} ( 3 y ^{2} - 4 ) ( 6 x ^{2} + 4 )}{( 3 y ^{2} - 4 ) ^{2}}

\frac{d}{d x} (6 x^{2} + 4) = 12 x

\frac{d}{d x} (3 y^{2} - 4) = 6 y \frac{d y}{d x}

= \frac{12 x ( 3 y ^{2} - 4 ) - 6 y \frac{d y}{d x} ( 6 x ^{2} + 4 )}{( 3 y ^{2} - 4 ) ^{2}}

\frac{d}{d x} ((4 x^{2} + 4) (2 x^{2} + 2 x))

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

\frac{d}{d x} (\frac{10}{x ^{2} + 1})

x \to - 4 lim (\frac{1}{2} x - 1)

\frac{7 e ^{x} x ^{\frac{5}{2}}}{2} + e^{x} x^{\frac{7}{2}}

What is the derivative of (6x^2+4/(3y^2-4)) ?
The derivative of (6x^2+4/(3y^2-4)) is (12x(3y^2-4)-6y\frac{dy)/(dx)(6x^2+4)}{(3y^2-4)^2}
What is the first derivative of (6x^2+4/(3y^2-4)) ?
The first derivative of (6x^2+4/(3y^2-4)) is (12x(3y^2-4)-6y\frac{dy)/(dx)(6x^2+4)}{(3y^2-4)^2}