What is the derivative of 1/(3x^4+1) ?

The derivative of 1/(3x^4+1) is -(12x^3)/((3x^4+1)^2)

What is the first derivative of 1/(3x^4+1) ?

The first derivative of 1/(3x^4+1) is -(12x^3)/((3x^4+1)^2)

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{4} + 1) = 12 x^{3}

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

Simplify

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

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

\int sec^{2} (2 x) - 1 d x

\int tan^{2} (x) sec^{10} (x) d x

x \to 0 + lim (\frac{tan ( x )}{x})

y^{^{''''}} + 50 y^{^{''}} + 625 y = 0

What is the derivative of 1/(3x^4+1) ?
The derivative of 1/(3x^4+1) is -(12x^3)/((3x^4+1)^2)
What is the first derivative of 1/(3x^4+1) ?
The first derivative of 1/(3x^4+1) is -(12x^3)/((3x^4+1)^2)