What is the derivative of 8/(3+x^2) ?

The derivative of 8/(3+x^2) is -(16x)/((3+x^2)^2)

What is the first derivative of 8/(3+x^2) ?

The first derivative of 8/(3+x^2) is -(16x)/((3+x^2)^2)

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\frac{d}{d x} (\frac{8}{3 + x ^{2}})

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

Solution steps

\frac{d}{d x} (\frac{8}{3 + x ^{2}})

Take the constant out:

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

= 8 \frac{d}{d x} (\frac{1}{3 + x ^{2}})

Apply exponent rule:

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

= 8 \frac{d}{d x} ((3 + x^{2})^{- 1})

Apply the chain rule:

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

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

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

= 8 (- \frac{1}{( 3 + x ^{2} ) ^{2}} \cdot 2 x)

Simplify

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

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

\int x^{\frac{1}{2}} d x

\frac{d}{d x} (\frac{y - 2 x}{2 y - x})

\frac{1}{1 + 2 x}

\int \frac{x ^{3} - 1}{x} d x

\frac{d y}{d x} = 21 x^{2} e^{- y}

What is the derivative of 8/(3+x^2) ?
The derivative of 8/(3+x^2) is -(16x)/((3+x^2)^2)
What is the first derivative of 8/(3+x^2) ?
The first derivative of 8/(3+x^2) is -(16x)/((3+x^2)^2)