What is the derivative of (24/(x^2+12)) ?

The derivative of (24/(x^2+12)) is -(48x)/((x^2+12)^2)

What is the first derivative of (24/(x^2+12)) ?

The first derivative of (24/(x^2+12)) is -(48x)/((x^2+12)^2)

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\frac{d}{d x} (\frac{24}{x ^{2} + 12})

- \frac{48 x}{( x ^{2} + 12 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{24}{x ^{2} + 12})

Take the constant out:

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

= 24 \frac{d}{d x} (\frac{1}{x ^{2} + 12})

Apply exponent rule:

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

= 24 \frac{d}{d x} ((x^{2} + 12)^{- 1})

Apply the chain rule:

- \frac{1}{( x ^{2} + 12 ) ^{2}} \frac{d}{d x} (x^{2} + 12)

= - \frac{1}{( x ^{2} + 12 ) ^{2}} \frac{d}{d x} (x^{2} + 12)

\frac{d}{d x} (x^{2} + 12) = 2 x

= 24 (- \frac{1}{( x ^{2} + 12 ) ^{2}} \cdot 2 x)

Simplify

24 (- \frac{1}{( x ^{2} + 12 ) ^{2}} \cdot 2 x) : - \frac{48 x}{( x ^{2} + 12 ) ^{2}}

= - \frac{48 x}{( x ^{2} + 12 ) ^{2}}

x \to 1 lim (e^{x} - 1)

\frac{d y}{d t} = \frac{y}{2} + 4 e^{\frac{t}{2}}

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

2 y^{^{''}} - 7 y^{^{'}} - 15 y = 0, y (0) = 2, y^{^{'}} (0) = 1

f (x) = x^{3} + 6 x^{2} + x df (x)

What is the derivative of (24/(x^2+12)) ?
The derivative of (24/(x^2+12)) is -(48x)/((x^2+12)^2)
What is the first derivative of (24/(x^2+12)) ?
The first derivative of (24/(x^2+12)) is -(48x)/((x^2+12)^2)