What is the derivative of-(128x/((16x^2+1)^2)) ?

The derivative of-(128x/((16x^2+1)^2)) is -(128(1-48x^2))/((16x^2+1)^3)

What is the first derivative of-(128x/((16x^2+1)^2)) ?

The first derivative of-(128x/((16x^2+1)^2)) is -(128(1-48x^2))/((16x^2+1)^3)

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\frac{d}{d x} (- \frac{128 x}{( 16 x ^{2} + 1 ) ^{2}})

- \frac{128 ( 1 - 48 x ^{2} )}{( 16 x ^{2} + 1 ) ^{3}}

Solution steps

\frac{d}{d x} (- \frac{128 x}{( 16 x ^{2} + 1 ) ^{2}})

Take the constant out:

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

= - 128 \frac{d}{d x} (\frac{x}{( 16 x ^{2} + 1 ) ^{2}})

Apply the Quotient Rule:

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

= - 128 \cdot \frac{\frac{d x}{d x} ( 16 x ^{2} + 1 ) ^{2} - \frac{d}{d x} ( ( 16 x ^{2} + 1 ) ^{2} ) x}{( ( 16 x ^{2} + 1 ) ^{2} ) ^{2}}

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

\frac{d}{d x} ((16 x^{2} + 1)^{2}) = 64 x (16 x^{2} + 1)

= - 128 \cdot \frac{1 \cdot ( 16 x ^{2} + 1 ) ^{2} - 64 x ( 16 x ^{2} + 1 ) x}{( ( 16 x ^{2} + 1 ) ^{2} ) ^{2}}

Simplify

- 128 \cdot \frac{1 \cdot ( 16 x ^{2} + 1 ) ^{2} - 64 x ( 16 x ^{2} + 1 ) x}{( ( 16 x ^{2} + 1 ) ^{2} ) ^{2}} : - \frac{128 ( 1 - 48 x ^{2} )}{( 16 x ^{2} + 1 ) ^{3}}

= - \frac{128 ( 1 - 48 x ^{2} )}{( 16 x ^{2} + 1 ) ^{3}}

f (t) = \frac{2 t}{1 + t ^{2}}

f (x) = 10 x^{4}

\frac{d}{d x} (9 x^{4} y^{8} - 2 x^{m} y^{5} - x^{9} y^{n})

\int_{0}^{1} e^{x^{5}} d x

n = 5 \sum \infty \frac{7 ^{n}}{1 0 ^{n}}

What is the derivative of-(128x/((16x^2+1)^2)) ?
The derivative of-(128x/((16x^2+1)^2)) is -(128(1-48x^2))/((16x^2+1)^3)
What is the first derivative of-(128x/((16x^2+1)^2)) ?
The first derivative of-(128x/((16x^2+1)^2)) is -(128(1-48x^2))/((16x^2+1)^3)