What is the derivative of (x-9/(8x^2e^x)) ?

The derivative of (x-9/(8x^2e^x)) is (-x^2+8x+18)/(8e^xx^3)

What is the first derivative of (x-9/(8x^2e^x)) ?

The first derivative of (x-9/(8x^2e^x)) is (-x^2+8x+18)/(8e^xx^3)

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

\frac{- x ^{2} + 8 x + 18}{8 e ^{x} x ^{3}}

Solution steps

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

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (\frac{x - 9}{x ^{2} e ^{x}})

Apply the Quotient Rule:

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

= \frac{1}{8} \cdot \frac{\frac{d}{d x} ( x - 9 ) x ^{2} e ^{x} - \frac{d}{d x} ( x ^{2} e ^{x} ) ( x - 9 )}{( x ^{2} e ^{x} ) ^{2}}

\frac{d}{d x} (x - 9) = 1

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

= \frac{1}{8} \cdot \frac{1 \cdot x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( x - 9 )}{( x ^{2} e ^{x} ) ^{2}}

Simplify

\frac{1}{8} \cdot \frac{1 \cdot x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( x - 9 )}{( x ^{2} e ^{x} ) ^{2}} : \frac{- x ^{2} + 8 x + 18}{8 e ^{x} x ^{3}}

= \frac{- x ^{2} + 8 x + 18}{8 e ^{x} x ^{3}}

\frac{d y}{d x} = 4 x (x - 2)

n = 1 \sum \infty - \frac{1}{n}

\int e^{3 x^{2}} d x

\frac{\partial}{\partial θ} (r sin (θ))

\int \frac{2}{x ^{3} + x ^{2} + x + 1} d x

What is the derivative of (x-9/(8x^2e^x)) ?
The derivative of (x-9/(8x^2e^x)) is (-x^2+8x+18)/(8e^xx^3)
What is the first derivative of (x-9/(8x^2e^x)) ?
The first derivative of (x-9/(8x^2e^x)) is (-x^2+8x+18)/(8e^xx^3)