What is (\partial)/(\partial x)((5(1-x))/(e^x)) ?

The answer to (\partial)/(\partial x)((5(1-x))/(e^x)) is -(5(-x+2))/(e^x)

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(\partial)/(\partial x)((5(1-x))/(e^x))

\frac{\partial}{\partial x} (\frac{5 ( 1 - x )}{e ^{x}})

- \frac{5 ( - x + 2 )}{e ^{x}}

Solution steps

\frac{\partial}{\partial x} (\frac{5 ( 1 - x )}{e ^{x}})

Take the constant out:

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

= 5 \frac{\partial}{\partial x} (\frac{1 - x}{e ^{x}})

Apply the Quotient Rule:

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

= 5 \cdot \frac{\frac{\partial}{\partial x} ( 1 - x ) e ^{x} - \frac{\partial}{\partial x} ( e ^{x} ) ( 1 - x )}{( e ^{x} ) ^{2}}

\frac{\partial}{\partial x} (1 - x) = - 1

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

= 5 \cdot \frac{( - 1 ) e ^{x} - e ^{x} ( 1 - x )}{( e ^{x} ) ^{2}}

Simplify

5 \cdot \frac{( - 1 ) e ^{x} - e ^{x} ( 1 - x )}{( e ^{x} ) ^{2}} : - \frac{5 ( - x + 2 )}{e ^{x}}

= - \frac{5 ( - x + 2 )}{e ^{x}}

\frac{d}{d x} (1 + 3 cos (2 x))

(ln (x^{2} - 1))^{^{'}}

y^{^{''}} - y^{^{'}} + 30 y = 0

3 sin (2 t) - 3 cos (2 t)

\frac{d}{d x} (x^{y} - y^{x})

What is (\partial)/(\partial x)((5(1-x))/(e^x)) ?
The answer to (\partial)/(\partial x)((5(1-x))/(e^x)) is -(5(-x+2))/(e^x)