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

The derivative of y=((2x-3)/((x^2+4)^2)) is (2(-3x^2+6x+4))/((x^2+4)^3)

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

The first derivative of y=((2x-3)/((x^2+4)^2)) is (2(-3x^2+6x+4))/((x^2+4)^3)

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

(1, 4), (4, 1)

n = 0 \sum \infty \frac{( n ! ) ^{2}}{2 ^{n^{2}}}

y = (2 x + e^{- x^{2}})^{3}

\frac{d P}{d t} = P (a - b P)

y = 2 x, y = 2 x^{2} - 4

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