What is the derivative of f(x)=(x^3+3x^2+1)^{-3} ?

The derivative of f(x)=(x^3+3x^2+1)^{-3} is -(3(3x^2+6x))/((x^3+3x^2+1)^4)

What is the first derivative of f(x)=(x^3+3x^2+1)^{-3} ?

The first derivative of f(x)=(x^3+3x^2+1)^{-3} is -(3(3x^2+6x))/((x^3+3x^2+1)^4)

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derivative of

f (x) = (x^{3} + 3 x^{2} + 1)^{- 3}

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

x \to \infty lim (2 - 1)

\int (x^{2} - 9) d x

\int \frac{x ( x - 6 )}{( x - 3 ) ^{3}} d x

\int \frac{3}{9 - 4 x ^{2}} d x

y^{^{''}} - y^{^{'}} - 6 y = - 4 e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

What is the derivative of f(x)=(x^3+3x^2+1)^{-3} ?
The derivative of f(x)=(x^3+3x^2+1)^{-3} is -(3(3x^2+6x))/((x^3+3x^2+1)^4)
What is the first derivative of f(x)=(x^3+3x^2+1)^{-3} ?
The first derivative of f(x)=(x^3+3x^2+1)^{-3} is -(3(3x^2+6x))/((x^3+3x^2+1)^4)