What is the derivative of (x^2e^x)/(x^2+e^x) ?

The derivative of (x^2e^x)/(x^2+e^x) is (e^xx^4+2e^{2x}x)/((x^2+e^x)^2)

What is the first derivative of (x^2e^x)/(x^2+e^x) ?

The first derivative of (x^2e^x)/(x^2+e^x) is (e^xx^4+2e^{2x}x)/((x^2+e^x)^2)

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derivative of

\frac{x ^{2} e ^{x}}{x ^{2} + e ^{x}}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(2 x e^{x} + e^{x} x^{2}) (x^{2} + e^{x}) - (2 x + e^{x}) x^{2} e^{x} : e^{x} x^{4} + 2 e^{2 x} x

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

y^{^{''}} + 25 y = 0

\frac{d}{d x} (x sec (\frac{1}{x}))

\int \frac{x ^{3}}{4 x ^{2} + 16} d x

\int \frac{e ^{- 3 x}}{1 - e ^{- 3 x}} d x

9

What is the derivative of (x^2e^x)/(x^2+e^x) ?
The derivative of (x^2e^x)/(x^2+e^x) is (e^xx^4+2e^{2x}x)/((x^2+e^x)^2)
What is the first derivative of (x^2e^x)/(x^2+e^x) ?
The first derivative of (x^2e^x)/(x^2+e^x) is (e^xx^4+2e^{2x}x)/((x^2+e^x)^2)