What is the derivative of y=(e^x)/(8x^4) ?

The derivative of y=(e^x)/(8x^4) is (e^x(x-4))/(8x^5)

What is the first derivative of y=(e^x)/(8x^4) ?

The first derivative of y=(e^x)/(8x^4) is (e^x(x-4))/(8x^5)

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derivative of

y = \frac{e ^{x}}{8 x ^{4}}

\frac{e ^{x} ( x - 4 )}{8 x ^{5}}

Solution steps

\frac{d}{d x} (\frac{e ^{x}}{8 x ^{4}})

Take the constant out:

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

= \frac{1}{8} \frac{d}{d x} (\frac{e ^{x}}{x ^{4}})

Apply the Quotient Rule:

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

= \frac{1}{8} \cdot \frac{\frac{d}{d x} ( e ^{x} ) x ^{4} - \frac{d}{d x} ( x ^{4} ) e ^{x}}{( x ^{4} ) ^{2}}

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

\frac{d}{d x} (x^{4}) = 4 x^{3}

= \frac{1}{8} \cdot \frac{e ^{x} x ^{4} - 4 x ^{3} e ^{x}}{( x ^{4} ) ^{2}}

Simplify

\frac{1}{8} \cdot \frac{e ^{x} x ^{4} - 4 x ^{3} e ^{x}}{( x ^{4} ) ^{2}} : \frac{e ^{x} ( x - 4 )}{8 x ^{5}}

= \frac{e ^{x} ( x - 4 )}{8 x ^{5}}

n \to \infty lim (\frac{n ^{2}}{2 n - 1} sin (\frac{1}{n}))

\frac{d}{d x} (ln (x^{2} + 5))

\frac{d}{d x} ((5 + 3 x) sin (x))

(y^{4} x) (\frac{d y}{d x}) = 1 + x, y (1) = 15

\frac{d y}{d x} = 3 - 2 y

What is the derivative of y=(e^x)/(8x^4) ?
The derivative of y=(e^x)/(8x^4) is (e^x(x-4))/(8x^5)
What is the first derivative of y=(e^x)/(8x^4) ?
The first derivative of y=(e^x)/(8x^4) is (e^x(x-4))/(8x^5)