What is (\partial)/(\partial x)(((2+t^2))/(3x-x^3)) ?

The answer to (\partial)/(\partial x)(((2+t^2))/(3x-x^3)) is -((2+t^2)(3-3x^2))/((3x-x^3)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(((2+t^2))/(3x-x^3))

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

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

Solution steps

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

Treat

t

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = e^{2 - x}

\int_{0}^{3} (9 - x^{2}) d x

\frac{d}{d a} (\frac{tan ^{2} ( a ) - sin ^{2} ( a )}{tan ^{2} ( a )})

\frac{d}{d x} (8 x y - \frac{1}{4} (x + y)^{4})

f (x) = x^{2} cos (x^{2})

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