What is the derivative of (3x-4/(2x^2+5)) ?

The derivative of (3x-4/(2x^2+5)) is (-6x^2+16x+15)/((2x^2+5)^2)

What is the first derivative of (3x-4/(2x^2+5)) ?

The first derivative of (3x-4/(2x^2+5)) is (-6x^2+16x+15)/((2x^2+5)^2)

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\frac{d}{d x} (\frac{3 x - 4}{2 x ^{2} + 5})

\frac{- 6 x ^{2} + 16 x + 15}{( 2 x ^{2} + 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{3 x - 4}{2 x ^{2} + 5})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 x - 4 ) ( 2 x ^{2} + 5 ) - \frac{d}{d x} ( 2 x ^{2} + 5 ) ( 3 x - 4 )}{( 2 x ^{2} + 5 ) ^{2}}

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

\frac{d}{d x} (2 x^{2} + 5) = 4 x

= \frac{3 ( 2 x ^{2} + 5 ) - 4 x ( 3 x - 4 )}{( 2 x ^{2} + 5 ) ^{2}}

Expand

3 (2 x^{2} + 5) - 4 x (3 x - 4) : - 6 x^{2} + 16 x + 15

= \frac{- 6 x ^{2} + 16 x + 15}{( 2 x ^{2} + 5 ) ^{2}}

\frac{d}{d x} (x^{2} e^{- 2 x})

x \to \frac{1}{2} lim (∣ 2 x + 3 ∣)

\frac{\partial}{\partial x} (x^{6} y^{8} + 7 x^{8} y)

- \frac{y}{x}

(3 x + 6 y) d x + (6 x - 8 y 3) d y = 0

What is the derivative of (3x-4/(2x^2+5)) ?
The derivative of (3x-4/(2x^2+5)) is (-6x^2+16x+15)/((2x^2+5)^2)
What is the first derivative of (3x-4/(2x^2+5)) ?
The first derivative of (3x-4/(2x^2+5)) is (-6x^2+16x+15)/((2x^2+5)^2)