What is the derivative of 6/((x^2+x+1^2)) ?

The derivative of 6/((x^2+x+1^2)) is -(12(2x+1))/((x^2+x+1)^3)

What is the first derivative of 6/((x^2+x+1^2)) ?

The first derivative of 6/((x^2+x+1^2)) is -(12(2x+1))/((x^2+x+1)^3)

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\frac{d}{d x} (\frac{6}{( x ^{2} + x + 1 ) ^{2}})

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

Solution steps

\frac{d}{d x} (\frac{6}{( x ^{2} + x + 1 ) ^{2}})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{1}{( x ^{2} + x + 1 ) ^{2}})

Apply exponent rule:

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

= 6 \frac{d}{d x} ((x^{2} + x + 1)^{- 2})

Apply the chain rule:

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

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

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

= 6 (- \frac{2}{( x ^{2} + x + 1 ) ^{3}} (2 x + 1))

Simplify

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

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

f (x) = 5 - x

x \to 3 - lim (ln (\frac{1}{3 - x}))

∣ x - 5 ∣ + 1

- 4 sin (3 t)

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

What is the derivative of 6/((x^2+x+1^2)) ?
The derivative of 6/((x^2+x+1^2)) is -(12(2x+1))/((x^2+x+1)^3)
What is the first derivative of 6/((x^2+x+1^2)) ?
The first derivative of 6/((x^2+x+1^2)) is -(12(2x+1))/((x^2+x+1)^3)