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

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

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

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

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derivative of

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

(x + 1) \frac{d y}{d x} = (x + 6)

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

\int \frac{1}{( 5 x + 2 ) ^{3}} d x

\frac{\partial}{\partial y} (x e^{5 x y})

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