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

The derivative of (3x+5/(x^2+4)) is (-3x^2-10x+12)/((x^2+4)^2)

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

The first derivative of (3x+5/(x^2+4)) is (-3x^2-10x+12)/((x^2+4)^2)

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

\frac{- 3 x ^{2} - 10 x + 12}{( x ^{2} + 4 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (3 x + 5) = 3

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

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

Expand

3 (x^{2} + 4) - 2 x (3 x + 5) : - 3 x^{2} - 10 x + 12

= \frac{- 3 x ^{2} - 10 x + 12}{( x ^{2} + 4 ) ^{2}}

x \to 8 lim (4 x)

n = 1 \sum \infty 6 (\frac{5}{4})^{n}

n = 0 \sum \infty \frac{( 1 ^{n} )}{n !}

\frac{\partial}{\partial x} (x^{2} e^{x})

\frac{d}{d x} ((1 - x) ln (1 - x))

What is the derivative of (3x+5/(x^2+4)) ?
The derivative of (3x+5/(x^2+4)) is (-3x^2-10x+12)/((x^2+4)^2)
What is the first derivative of (3x+5/(x^2+4)) ?
The first derivative of (3x+5/(x^2+4)) is (-3x^2-10x+12)/((x^2+4)^2)