What is the derivative of (sin(x)/(8xe^x)) ?

The derivative of (sin(x)/(8xe^x)) is (e^xxcos(x)-sin(x)(e^x+e^xx))/(8e^{2x)x^2}

What is the first derivative of (sin(x)/(8xe^x)) ?

The first derivative of (sin(x)/(8xe^x)) is (e^xxcos(x)-sin(x)(e^x+e^xx))/(8e^{2x)x^2}

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\frac{d}{d x} (\frac{sin ( x )}{8 x e ^{x}})

\frac{e ^{x} x cos ( x ) - sin ( x ) ( e ^{x} + e ^{x} x )}{8 e ^{2 x} x ^{2}}

Solution steps

\frac{d}{d x} (\frac{sin ( x )}{8 x e ^{x}})

Take the constant out:

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

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

\frac{d}{d x} (sin (x)) = cos (x)

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

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

Simplify

\frac{1}{8} \cdot \frac{cos ( x ) x e ^{x} - ( e ^{x} + e ^{x} x ) sin ( x )}{( x e ^{x} ) ^{2}} : \frac{e ^{x} x cos ( x ) - sin ( x ) ( e ^{x} + e ^{x} x )}{8 e ^{2 x} x ^{2}}

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

What is the derivative of (sin(x)/(8xe^x)) ?
The derivative of (sin(x)/(8xe^x)) is (e^xxcos(x)-sin(x)(e^x+e^xx))/(8e^{2x)x^2}
What is the first derivative of (sin(x)/(8xe^x)) ?
The first derivative of (sin(x)/(8xe^x)) is (e^xxcos(x)-sin(x)(e^x+e^xx))/(8e^{2x)x^2}