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

The derivative of 3/(x^4+2) is -(12x^3)/((x^4+2)^2)

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

The first derivative of 3/(x^4+2) is -(12x^3)/((x^4+2)^2)

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

- \frac{12 x ^{3}}{( x ^{4} + 2 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

3 (- \frac{1}{( x ^{4} + 2 ) ^{2}} \cdot 4 x^{3}) : - \frac{12 x ^{3}}{( x ^{4} + 2 ) ^{2}}

= - \frac{12 x ^{3}}{( x ^{4} + 2 ) ^{2}}

\int_{0}^{2} 6 x^{2} - 18 x + 12 d x

f (x) = x^{2} - \frac{1}{x}, (2, 3.5)

n = 1 \sum \infty \frac{( ( - 7 ) ^{n - 1} )}{8 ^{n}}

y^{^{''}} - 14 y^{^{'}} + 49 y = t^{- 2} e^{7 t}

x \to - \infty lim (\frac{7}{x} - 4)

What is the derivative of 3/(x^4+2) ?
The derivative of 3/(x^4+2) is -(12x^3)/((x^4+2)^2)
What is the first derivative of 3/(x^4+2) ?
The first derivative of 3/(x^4+2) is -(12x^3)/((x^4+2)^2)