What is the derivative of 6/((1+x^4)) ?

The derivative of 6/((1+x^4)) is -(24)/((1+x)^5)

What is the first derivative of 6/((1+x^4)) ?

The first derivative of 6/((1+x^4)) is -(24)/((1+x)^5)

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\frac{d}{d x} (\frac{6}{( 1 + x ) ^{4}})

- \frac{24}{( 1 + x ) ^{5}}

Solution steps

\frac{d}{d x} (\frac{6}{( 1 + x ) ^{4}})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{1}{( 1 + x ) ^{4}})

Apply exponent rule:

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

= 6 \frac{d}{d x} ((1 + x)^{- 4})

Apply the chain rule:

- \frac{4}{( 1 + x ) ^{5}} \frac{d}{d x} (1 + x)

= - \frac{4}{( 1 + x ) ^{5}} \frac{d}{d x} (1 + x)

\frac{d}{d x} (1 + x) = 1

= 6 (- \frac{4}{( 1 + x ) ^{5}} \cdot 1)

Simplify

6 (- \frac{4}{( 1 + x ) ^{5}} \cdot 1) : - \frac{24}{( 1 + x ) ^{5}}

= - \frac{24}{( 1 + x ) ^{5}}

\frac{d}{d x} (\frac{6 x ^{3} - 2}{x ^{2} - 4})

f (x) = x^{2} + x

\int sec (5 x) tan (5 x) d x

\frac{d}{d x} (sec (\frac{1}{x}))

\int \frac{1}{( x ^{2} + 9 ) ^{\frac{5}{2}}} d x

What is the derivative of 6/((1+x^4)) ?
The derivative of 6/((1+x^4)) is -(24)/((1+x)^5)
What is the first derivative of 6/((1+x^4)) ?
The first derivative of 6/((1+x^4)) is -(24)/((1+x)^5)