What is the derivative of (1x^3)/(2x^4) ?

The derivative of (1x^3)/(2x^4) is -1/(2x^2)

What is the first derivative of (1x^3)/(2x^4) ?

The first derivative of (1x^3)/(2x^4) is -1/(2x^2)

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derivative of

\frac{1 x ^{3}}{2 x ^{4}}

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

Solution steps

\frac{d}{d x} (\frac{1 \cdot x ^{3}}{2 x ^{4}})

Take the constant out:

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

= \frac{1}{2} \frac{d}{d x} (\frac{x ^{3}}{x ^{4}})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{3}) = 3 x^{2}

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

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

Simplify

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

= - \frac{1}{2 x ^{2}}

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

\frac{\partial}{\partial x} (3 x y^{2} - 7)

\int \frac{2}{3} sec^{2} (π x) d x

\int \frac{cos ( x ) - 1}{x} d x

(8.6) (3.6)

What is the derivative of (1x^3)/(2x^4) ?
The derivative of (1x^3)/(2x^4) is -1/(2x^2)
What is the first derivative of (1x^3)/(2x^4) ?
The first derivative of (1x^3)/(2x^4) is -1/(2x^2)