What is the derivative of (x^9e^x/(x^9+e^x)) ?

The derivative of (x^9e^x/(x^9+e^x)) is (e^xx^{18}+9e^{2x}x^8)/((x^9+e^x)^2)

What is the first derivative of (x^9e^x/(x^9+e^x)) ?

The first derivative of (x^9e^x/(x^9+e^x)) is (e^xx^{18}+9e^{2x}x^8)/((x^9+e^x)^2)

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\frac{d}{d x} (\frac{x ^{9} e ^{x}}{x ^{9} + e ^{x}})

\frac{e ^{x} x ^{18} + 9 e ^{2 x} x ^{8}}{( x ^{9} + e ^{x} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{9} e ^{x}}{x ^{9} + e ^{x}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{9} e ^{x} ) ( x ^{9} + e ^{x} ) - \frac{d}{d x} ( x ^{9} + e ^{x} ) x ^{9} e ^{x}}{( x ^{9} + e ^{x} ) ^{2}}

\frac{d}{d x} (x^{9} e^{x}) = 9 x^{8} e^{x} + e^{x} x^{9}

\frac{d}{d x} (x^{9} + e^{x}) = 9 x^{8} + e^{x}

= \frac{( 9 x ^{8} e ^{x} + e ^{x} x ^{9} ) ( x ^{9} + e ^{x} ) - ( 9 x ^{8} + e ^{x} ) x ^{9} e ^{x}}{( x ^{9} + e ^{x} ) ^{2}}

Expand

(9 x^{8} e^{x} + e^{x} x^{9}) (x^{9} + e^{x}) - (9 x^{8} + e^{x}) x^{9} e^{x} : e^{x} x^{18} + 9 e^{2 x} x^{8}

= \frac{e ^{x} x ^{18} + 9 e ^{2 x} x ^{8}}{( x ^{9} + e ^{x} ) ^{2}}

n = 1 \sum \infty 2 n e^{- n^{2}}

(y^{3} - 2 x^{2} y) d x + 2 x^{3} d y = 0

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

cos (θ)

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

What is the derivative of (x^9e^x/(x^9+e^x)) ?
The derivative of (x^9e^x/(x^9+e^x)) is (e^xx^{18}+9e^{2x}x^8)/((x^9+e^x)^2)
What is the first derivative of (x^9e^x/(x^9+e^x)) ?
The first derivative of (x^9e^x/(x^9+e^x)) is (e^xx^{18}+9e^{2x}x^8)/((x^9+e^x)^2)