What is the derivative of f(x)=(e^x)/(4-4x) ?

The derivative of f(x)=(e^x)/(4-4x) is (e^x(-x+2))/(4(x-1)^2)

What is the first derivative of f(x)=(e^x)/(4-4x) ?

The first derivative of f(x)=(e^x)/(4-4x) is (e^x(-x+2))/(4(x-1)^2)

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (4 - 4 x) = - 4

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

Simplify

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

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

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

f (x) = 6 - x^{2}, g (x) = x

1 - 3 x

\int sin^{2} (r) d r

\int (6 csc (x) cot (x)) d x

What is the derivative of f(x)=(e^x)/(4-4x) ?
The derivative of f(x)=(e^x)/(4-4x) is (e^x(-x+2))/(4(x-1)^2)
What is the first derivative of f(x)=(e^x)/(4-4x) ?
The first derivative of f(x)=(e^x)/(4-4x) is (e^x(-x+2))/(4(x-1)^2)