What is the derivative of (e^x/(4x^2)) ?

The derivative of (e^x/(4x^2)) is (e^x(x-2))/(4x^3)

What is the first derivative of (e^x/(4x^2)) ?

The first derivative of (e^x/(4x^2)) is (e^x(x-2))/(4x^3)

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

\frac{e ^{x} ( x - 2 )}{4 x ^{3}}

Solution steps

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

Take the constant out:

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

= \frac{1}{4} \frac{d}{d x} (\frac{e ^{x}}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{1}{4} \cdot \frac{\frac{d}{d x} ( e ^{x} ) x ^{2} - \frac{d}{d x} ( x ^{2} ) e ^{x}}{( x ^{2} ) ^{2}}

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

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

= \frac{1}{4} \cdot \frac{e ^{x} x ^{2} - 2 x e ^{x}}{( x ^{2} ) ^{2}}

Simplify

\frac{1}{4} \cdot \frac{e ^{x} x ^{2} - 2 x e ^{x}}{( x ^{2} ) ^{2}} : \frac{e ^{x} ( x - 2 )}{4 x ^{3}}

= \frac{e ^{x} ( x - 2 )}{4 x ^{3}}

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

y^{^{''}} - 8 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 9

\frac{\partial}{\partial x} (x^{- 3})

x \to - 1 lim (4 x^{2} - 8 x + 3)

\int e^{- x} cos (3 x) d x

What is the derivative of (e^x/(4x^2)) ?
The derivative of (e^x/(4x^2)) is (e^x(x-2))/(4x^3)
What is the first derivative of (e^x/(4x^2)) ?
The first derivative of (e^x/(4x^2)) is (e^x(x-2))/(4x^3)