What is the derivative of (64/(x^2+16)) ?

The derivative of (64/(x^2+16)) is -(128x)/((x^2+16)^2)

What is the first derivative of (64/(x^2+16)) ?

The first derivative of (64/(x^2+16)) is -(128x)/((x^2+16)^2)

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

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

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int \frac{x ^{2} - a ^{2}}{x ^{2}} d x

\int \frac{1}{2} cos (x) d x

y \to 2 lim (\frac{y ^{3} - 6 y ^{2} + 12 y - 8}{y ^{4} - 4 y ^{3} + 16 y - 16})

\frac{d y}{d x} = x - x y

x \to 4 + lim (x^{2} - 16)

What is the derivative of (64/(x^2+16)) ?
The derivative of (64/(x^2+16)) is -(128x)/((x^2+16)^2)
What is the first derivative of (64/(x^2+16)) ?
The first derivative of (64/(x^2+16)) is -(128x)/((x^2+16)^2)