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

The derivative of (xe^x-9x/(e^x)) is (e^x-9+9x)/(e^x)

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

The first derivative of (xe^x-9x/(e^x)) is (e^x-9+9x)/(e^x)

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

\frac{e ^{x} - 9 + 9 x}{e ^{x}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

= \frac{e ^{x} - 9 + 9 x}{e ^{x}}

9 x^{- 4} - 9 x^{- 1}

\int_{- \infty}^{1} 5^{x} d x

x^{^{''}} + 9 x - 27 = 0

\int_{3}^{5} \frac{2}{4 + ( x - 3 ) ^{2}} d x

\frac{d}{d x} ((2 x - 1) (3 - 4 x))

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