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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derivative of

\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}}

f (x) = x^{2} - 4 x, at x = 1

\frac{d}{d x} (- 5 sin (x))

\frac{\partial}{\partial x} (2 - x^{2} - y^{2})

\frac{d}{d x} (- e^{- 4 x})

\int_{3}^{\infty} \frac{e ^{- \frac{1}{x}}}{x ^{2}} d x

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)