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

The derivative of 1/(x^3+4x^2) is -(3x+8)/(x^3(x+4)^2)

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

The first derivative of 1/(x^3+4x^2) is -(3x+8)/(x^3(x+4)^2)

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

- \frac{3 x + 8}{x ^{3} ( x + 4 ) ^{2}}

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (x^{3} + 4 x^{2}) = 3 x^{2} + 8 x

= - \frac{1}{( x ^{3} + 4 x ^{2} ) ^{2}} (3 x^{2} + 8 x)

Simplify

- \frac{1}{( x ^{3} + 4 x ^{2} ) ^{2}} (3 x^{2} + 8 x) : - \frac{3 x + 8}{x ^{3} ( x + 4 ) ^{2}}

= - \frac{3 x + 8}{x ^{3} ( x + 4 ) ^{2}}

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

\int 7 tan^{2} (x) + tan^{4} (x) d x

\int 3 x^{3} - 5 x^{2} + 7 x + 2 d x

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

x \to \infty lim (\frac{3 x ^{2} + 2 x - 8}{4 x ^{4} - 5 x + 4})

What is the derivative of 1/(x^3+4x^2) ?
The derivative of 1/(x^3+4x^2) is -(3x+8)/(x^3(x+4)^2)
What is the first derivative of 1/(x^3+4x^2) ?
The first derivative of 1/(x^3+4x^2) is -(3x+8)/(x^3(x+4)^2)