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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

\frac{d}{d x} (1 - x) = - 1

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

Simplify

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

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

\frac{\partial}{\partial x} (ln (y))

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

\frac{d}{d x} (x cot (x))

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

\int ln (x) \cdot sin (x) d x

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