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

The derivative of 4/(x^2+2) is -(8x)/((x^2+2)^2)

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

The first derivative of 4/(x^2+2) is -(8x)/((x^2+2)^2)

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

- \frac{8 x}{( x ^{2} + 2 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{0}^{l n (2)} e^{2 x} d x

\int \frac{9}{2} (t + 1)^{\frac{1}{2}} d t

y^{^{'}} = x^{3} (1 - y), y (0) = 3

\int \frac{x - 1}{x + 1} d x

e^{- 1}

What is the derivative of 4/(x^2+2) ?
The derivative of 4/(x^2+2) is -(8x)/((x^2+2)^2)
What is the first derivative of 4/(x^2+2) ?
The first derivative of 4/(x^2+2) is -(8x)/((x^2+2)^2)