What is the derivative of f(x)=(x-7)/(9x^2e^x) ?

The derivative of f(x)=(x-7)/(9x^2e^x) is (-x^2+6x+14)/(9e^xx^3)

What is the first derivative of f(x)=(x-7)/(9x^2e^x) ?

The first derivative of f(x)=(x-7)/(9x^2e^x) is (-x^2+6x+14)/(9e^xx^3)

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derivative of

f (x) = \frac{x - 7}{9 x ^{2} e ^{x}}

\frac{- x ^{2} + 6 x + 14}{9 e ^{x} x ^{3}}

Solution steps

\frac{d}{d x} (\frac{x - 7}{9 x ^{2} e ^{x}})

Take the constant out:

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

= \frac{1}{9} \frac{d}{d x} (\frac{x - 7}{x ^{2} e ^{x}})

Apply the Quotient Rule:

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

= \frac{1}{9} \cdot \frac{\frac{d}{d x} ( x - 7 ) x ^{2} e ^{x} - \frac{d}{d x} ( x ^{2} e ^{x} ) ( x - 7 )}{( x ^{2} e ^{x} ) ^{2}}

\frac{d}{d x} (x - 7) = 1

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

= \frac{1}{9} \cdot \frac{1 \cdot x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( x - 7 )}{( x ^{2} e ^{x} ) ^{2}}

Simplify

\frac{1}{9} \cdot \frac{1 \cdot x ^{2} e ^{x} - ( 2 x e ^{x} + e ^{x} x ^{2} ) ( x - 7 )}{( x ^{2} e ^{x} ) ^{2}} : \frac{- x ^{2} + 6 x + 14}{9 e ^{x} x ^{3}}

= \frac{- x ^{2} + 6 x + 14}{9 e ^{x} x ^{3}}

\frac{\partial}{\partial x} ((e^{x} - 3) sin (y))

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

\frac{d}{d x} ((x^{3} - 4 x) \cdot e^{3 x})

y^{^{'''}} - 8 = 0

What is the derivative of f(x)=(x-7)/(9x^2e^x) ?
The derivative of f(x)=(x-7)/(9x^2e^x) is (-x^2+6x+14)/(9e^xx^3)
What is the first derivative of f(x)=(x-7)/(9x^2e^x) ?
The first derivative of f(x)=(x-7)/(9x^2e^x) is (-x^2+6x+14)/(9e^xx^3)