What is the derivative of (xcos(x)/(1+e^x)) ?

The derivative of (xcos(x)/(1+e^x)) is ((cos(x)-xsin(x))(1+e^x)-e^xxcos(x))/((1+e^x)^2)

What is the first derivative of (xcos(x)/(1+e^x)) ?

The first derivative of (xcos(x)/(1+e^x)) is ((cos(x)-xsin(x))(1+e^x)-e^xxcos(x))/((1+e^x)^2)

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\frac{d}{d x} (\frac{x cos ( x )}{1 + e ^{x}})

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

Solution steps

\frac{d}{d x} (\frac{x cos ( x )}{1 + e ^{x}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x cos ( x ) ) ( 1 + e ^{x} ) - \frac{d}{d x} ( 1 + e ^{x} ) x cos ( x )}{( 1 + e ^{x} ) ^{2}}

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

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

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

\frac{\partial}{\partial x} (y^{5} sin (2 x))

4 - x^{2}, x^{2} - 2 x

y^{^{''}} - 2 y^{^{'}} + y = t e^{t} + 4

y^{^{''}} + 36 y = sec^{2} (6 x)

y^{^{'}} = 3 x^{2} (1 + y^{2})

What is the derivative of (xcos(x)/(1+e^x)) ?
The derivative of (xcos(x)/(1+e^x)) is ((cos(x)-xsin(x))(1+e^x)-e^xxcos(x))/((1+e^x)^2)
What is the first derivative of (xcos(x)/(1+e^x)) ?
The first derivative of (xcos(x)/(1+e^x)) is ((cos(x)-xsin(x))(1+e^x)-e^xxcos(x))/((1+e^x)^2)