What is the derivative of 1/(3+(2x+1^2)) ?

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 + (2 x + 1)^{2}) = 8 x + 4

= - \frac{1}{( 3 + ( 2 x + 1 ) ^{2} ) ^{2}} (8 x + 4)

Simplify

- \frac{1}{( 3 + ( 2 x + 1 ) ^{2} ) ^{2}} (8 x + 4) : - \frac{2 x + 1}{4 ( x ^{2} + x + 1 ) ^{2}}

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

cos (t) y^{^{'}} + sin (t) y - 1 = 0

f (x) = 2 x, (1, 2)

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

x \to 1 + lim (arcsin (x))

\frac{d}{d x} (ln (ln (ln (2 x))))

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