What is the derivative of f(x)=(e^{4x})/(x^{12)} ?

The derivative of f(x)=(e^{4x})/(x^{12)} is (4e^{4x}(x-3))/(x^{13)}

What is the first derivative of f(x)=(e^{4x})/(x^{12)} ?

The first derivative of f(x)=(e^{4x})/(x^{12)} is (4e^{4x}(x-3))/(x^{13)}

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derivative of

f (x) = \frac{e ^{4 x}}{x ^{12}}

\frac{4 e ^{4 x} ( x - 3 )}{x ^{13}}

Solution steps

\frac{d}{d x} (\frac{e ^{4 x}}{x ^{12}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( e ^{4 x} ) x ^{12} - \frac{d}{d x} ( x ^{12} ) e ^{4 x}}{( x ^{12} ) ^{2}}

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

\frac{d}{d x} (x^{12}) = 12 x^{11}

= \frac{e ^{4 x} \cdot 4 x ^{12} - 12 x ^{11} e ^{4 x}}{( x ^{12} ) ^{2}}

Simplify

\frac{e ^{4 x} \cdot 4 x ^{12} - 12 x ^{11} e ^{4 x}}{( x ^{12} ) ^{2}} : \frac{4 e ^{4 x} ( x - 3 )}{x ^{13}}

= \frac{4 e ^{4 x} ( x - 3 )}{x ^{13}}

\int_{0}^{1} t^{3} d t

\int (x + 4) \cdot e^{x^{2} + 8 x} d x

x^{3} - 8

x (7^{- 3 x})

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

What is the derivative of f(x)=(e^{4x})/(x^{12)} ?
The derivative of f(x)=(e^{4x})/(x^{12)} is (4e^{4x}(x-3))/(x^{13)}
What is the first derivative of f(x)=(e^{4x})/(x^{12)} ?
The first derivative of f(x)=(e^{4x})/(x^{12)} is (4e^{4x}(x-3))/(x^{13)}