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

The derivative of 1/(-3+(3x+1^2)) is -(18x+6)/((9x^2+6x-2)^2)

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

The first derivative of 1/(-3+(3x+1^2)) is -(18x+6)/((9x^2+6x-2)^2)

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

- \frac{18 x + 6}{( 9 x ^{2} + 6 x - 2 ) ^{2}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (- 3 + (3 x + 1)^{2}) = 18 x + 6

= - \frac{1}{( - 3 + ( 3 x + 1 ) ^{2} ) ^{2}} (18 x + 6)

Simplify

- \frac{1}{( - 3 + ( 3 x + 1 ) ^{2} ) ^{2}} (18 x + 6) : - \frac{18 x + 6}{( 9 x ^{2} + 6 x - 2 ) ^{2}}

= - \frac{18 x + 6}{( 9 x ^{2} + 6 x - 2 ) ^{2}}

y = sec (x), (\frac{π}{6}, 2 \frac{3}{3})

\frac{d y}{d x} = 1 + 8 y

tan^{8} (x)

\frac{\partial}{\partial y} (x^{3} + 2 x y + y^{2})

\frac{\partial}{\partial y} (3 x^{2} y^{2} + 3)

What is the derivative of 1/(-3+(3x+1^2)) ?
The derivative of 1/(-3+(3x+1^2)) is -(18x+6)/((9x^2+6x-2)^2)
What is the first derivative of 1/(-3+(3x+1^2)) ?
The first derivative of 1/(-3+(3x+1^2)) is -(18x+6)/((9x^2+6x-2)^2)