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

The derivative of 1/((x+1^3(3x-1)^2)) is -(3(5x+1))/((x+1)^4(3x-1)^3)

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

The first derivative of 1/((x+1^3(3x-1)^2)) is -(3(5x+1))/((x+1)^4(3x-1)^3)

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

- \frac{3 ( 5 x + 1 )}{( x + 1 ) ^{4} ( 3 x - 1 ) ^{3}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

- \frac{1}{( ( x + 1 ) ^{3} ( 3 x - 1 ) ^{2} ) ^{2}} (3 (x + 1)^{2} (3 x - 1)^{2} + 6 (3 x - 1) (x + 1)^{3}) : - \frac{3 ( 5 x + 1 )}{( x + 1 ) ^{4} ( 3 x - 1 ) ^{3}}

= - \frac{3 ( 5 x + 1 )}{( x + 1 ) ^{4} ( 3 x - 1 ) ^{3}}

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

y (θ) = \frac{π}{7} cos (θ)

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

\int e^{x} sin (y) d x

\frac{d}{d x} (y^{2} - 2 y \cdot cos (x))

What is the derivative of 1/((x+1^3(3x-1)^2)) ?
The derivative of 1/((x+1^3(3x-1)^2)) is -(3(5x+1))/((x+1)^4(3x-1)^3)
What is the first derivative of 1/((x+1^3(3x-1)^2)) ?
The first derivative of 1/((x+1^3(3x-1)^2)) is -(3(5x+1))/((x+1)^4(3x-1)^3)