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

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

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derivative of

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

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

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

\int \frac{5 x + 8}{x ^{2} + 4 x + 4} d x

x \to \infty lim ((1 + \frac{1}{2 x})^{3 x})

\int_{2}^{6} t^{3} ln (4 t) d t

y^{^{'}} = 5 y sin (4 x), y (\frac{π}{8}) = 3

n = 1 \sum \infty \frac{1}{π ^{n}}

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