What is the derivative of-(8x/((4x^2+1)^2)) ?

The derivative of-(8x/((4x^2+1)^2)) is -(8(1-12x^2))/((4x^2+1)^3)

What is the first derivative of-(8x/((4x^2+1)^2)) ?

The first derivative of-(8x/((4x^2+1)^2)) is -(8(1-12x^2))/((4x^2+1)^3)

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\frac{d}{d x} (- \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}})

- \frac{8 ( 1 - 12 x ^{2} )}{( 4 x ^{2} + 1 ) ^{3}}

Solution steps

\frac{d}{d x} (- \frac{8 x}{( 4 x ^{2} + 1 ) ^{2}})

Take the constant out:

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

= - 8 \frac{d}{d x} (\frac{x}{( 4 x ^{2} + 1 ) ^{2}})

Apply the Quotient Rule:

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

= - 8 \cdot \frac{\frac{d x}{d x} ( 4 x ^{2} + 1 ) ^{2} - \frac{d}{d x} ( ( 4 x ^{2} + 1 ) ^{2} ) x}{( ( 4 x ^{2} + 1 ) ^{2} ) ^{2}}

\frac{d x}{d x} = 1

\frac{d}{d x} ((4 x^{2} + 1)^{2}) = 16 x (4 x^{2} + 1)

= - 8 \cdot \frac{1 \cdot ( 4 x ^{2} + 1 ) ^{2} - 16 x ( 4 x ^{2} + 1 ) x}{( ( 4 x ^{2} + 1 ) ^{2} ) ^{2}}

Simplify

- 8 \cdot \frac{1 \cdot ( 4 x ^{2} + 1 ) ^{2} - 16 x ( 4 x ^{2} + 1 ) x}{( ( 4 x ^{2} + 1 ) ^{2} ) ^{2}} : - \frac{8 ( 1 - 12 x ^{2} )}{( 4 x ^{2} + 1 ) ^{3}}

= - \frac{8 ( 1 - 12 x ^{2} )}{( 4 x ^{2} + 1 ) ^{3}}

\frac{d}{d x} (arcsin (coth (x)))

ln (7 x)

y = 6 - x^{2}, y = 26 - 9 x

y^{^{'''}} + y = 0

What is the derivative of-(8x/((4x^2+1)^2)) ?
The derivative of-(8x/((4x^2+1)^2)) is -(8(1-12x^2))/((4x^2+1)^3)
What is the first derivative of-(8x/((4x^2+1)^2)) ?
The first derivative of-(8x/((4x^2+1)^2)) is -(8(1-12x^2))/((4x^2+1)^3)