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

The derivative of f(x)=(e^x)/(x^n) is (e^x(x-n))/(x^{1+n)}

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

The first derivative of f(x)=(e^x)/(x^n) is (e^x(x-n))/(x^{1+n)}

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derivative of

f (x) = \frac{e ^{x}}{x ^{n}}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x^{n}) = n x^{n - 1}

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

Simplify

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

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

\int_{- 1}^{1} \frac{x}{1 - x ^{2}} d x

x x + 5

f (x) = ln (\frac{x + 1}{x - 1})

\int x^{1} (2 + 3 x^{2})^{3} d x

(5 - 1) (5 - 4)

What is the derivative of f(x)=(e^x)/(x^n) ?
The derivative of f(x)=(e^x)/(x^n) is (e^x(x-n))/(x^{1+n)}
What is the first derivative of f(x)=(e^x)/(x^n) ?
The first derivative of f(x)=(e^x)/(x^n) is (e^x(x-n))/(x^{1+n)}