What is the derivative of ((x+4)(x^2-4x+16))/(x^3) ?

The derivative of ((x+4)(x^2-4x+16))/(x^3) is -(192)/(x^4)

What is the first derivative of ((x+4)(x^2-4x+16))/(x^3) ?

The first derivative of ((x+4)(x^2-4x+16))/(x^3) is -(192)/(x^4)

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derivative of

\frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}}

- \frac{192}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{( x + 4 ) ( x ^{2} - 4 x + 16 )}{x ^{3}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( ( x + 4 ) ( x ^{2} - 4 x + 16 ) ) x ^{3} - \frac{d}{d x} ( x ^{3} ) ( x + 4 ) ( x ^{2} - 4 x + 16 )}{( x ^{3} ) ^{2}}

\frac{d}{d x} ((x + 4) (x^{2} - 4 x + 16)) = 3 x^{2}

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

= \frac{3 x ^{2} x ^{3} - 3 x ^{2} ( x + 4 ) ( x ^{2} - 4 x + 16 )}{( x ^{3} ) ^{2}}

Simplify

\frac{3 x ^{2} x ^{3} - 3 x ^{2} ( x + 4 ) ( x ^{2} - 4 x + 16 )}{( x ^{3} ) ^{2}} : - \frac{192}{x ^{4}}

= - \frac{192}{x ^{4}}

\int \frac{1}{3} e^{2 - x} d x

\frac{d}{d x} (x 36 - x^{2})

\frac{d}{d x} ((2 x + 3) (5 x^{2} - 4 x))

\int_{0}^{b} e^{\frac{x}{2}} d x

\int \frac{x ^{3}}{3 x} d x

What is the derivative of ((x+4)(x^2-4x+16))/(x^3) ?
The derivative of ((x+4)(x^2-4x+16))/(x^3) is -(192)/(x^4)
What is the first derivative of ((x+4)(x^2-4x+16))/(x^3) ?
The first derivative of ((x+4)(x^2-4x+16))/(x^3) is -(192)/(x^4)