What is (\partial)/(\partial x)(1/(x^3+4x^2)) ?

The answer to (\partial)/(\partial x)(1/(x^3+4x^2)) is -(3x+8)/(x^3(x+4)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(1/(x^3+4x^2))

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{\partial}{\partial 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} (cot (8 x))

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

\int \frac{sec ( x ) tan ( x )}{sec ( x )} d x

\frac{2 - 5 x}{3 x ^{\frac{1}{3}}}

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

What is (\partial)/(\partial x)(1/(x^3+4x^2)) ?
The answer to (\partial)/(\partial x)(1/(x^3+4x^2)) is -(3x+8)/(x^3(x+4)^2)