What is the derivative of 3/(4+x^2) ?

The derivative of 3/(4+x^2) is -(6x)/((4+x^2)^2)

What is the first derivative of 3/(4+x^2) ?

The first derivative of 3/(4+x^2) is -(6x)/((4+x^2)^2)

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\frac{d}{d x} (\frac{3}{4 + x ^{2}})

- \frac{6 x}{( 4 + x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{3}{4 + x ^{2}})

Take the constant out:

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

= 3 \frac{d}{d x} (\frac{1}{4 + x ^{2}})

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

3 (- \frac{1}{( 4 + x ^{2} ) ^{2}} \cdot 2 x) : - \frac{6 x}{( 4 + x ^{2} ) ^{2}}

= - \frac{6 x}{( 4 + x ^{2} ) ^{2}}

in v erse l a pl a ce \frac{( s - 2 )}{s ^{2} ( s - 1 ) ^{2}}

sin (t) cos (x) d t - sin (x) cos (t) d x = 0

a re a y^{2} = x, x = 2 y

n = 1 \sum \infty \frac{n}{n + n ^{2}}

l a pl a ce t r an s f or m 2 e^{t}

What is the derivative of 3/(4+x^2) ?
The derivative of 3/(4+x^2) is -(6x)/((4+x^2)^2)
What is the first derivative of 3/(4+x^2) ?
The first derivative of 3/(4+x^2) is -(6x)/((4+x^2)^2)