What is the derivative of (x^4-5x^3+8/(x^2)) ?

The derivative of (x^4-5x^3+8/(x^2)) is (2x^4-5x^3-16)/(x^3)

What is the first derivative of (x^4-5x^3+8/(x^2)) ?

The first derivative of (x^4-5x^3+8/(x^2)) is (2x^4-5x^3-16)/(x^3)

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\frac{d}{d x} (\frac{x ^{4} - 5 x ^{3} + 8}{x ^{2}})

\frac{2 x ^{4} - 5 x ^{3} - 16}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{x ^{4} - 5 x ^{3} + 8}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{4} - 5 x ^{3} + 8 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( x ^{4} - 5 x ^{3} + 8 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (x^{4} - 5 x^{3} + 8) = 4 x^{3} - 15 x^{2}

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

= \frac{( 4 x ^{3} - 15 x ^{2} ) x ^{2} - 2 x ( x ^{4} - 5 x ^{3} + 8 )}{( x ^{2} ) ^{2}}

Simplify

\frac{( 4 x ^{3} - 15 x ^{2} ) x ^{2} - 2 x ( x ^{4} - 5 x ^{3} + 8 )}{( x ^{2} ) ^{2}} : \frac{2 x ^{4} - 5 x ^{3} - 16}{x ^{3}}

= \frac{2 x ^{4} - 5 x ^{3} - 16}{x ^{3}}

\frac{d}{d x} (e^{- x^{9}})

\int (tan^{2} (x) sec (x)) d x

x \to 1 lim (4 x + 3)

\int sin (x) cot (x) d x

x \to 0 - lim (ln (- x))

What is the derivative of (x^4-5x^3+8/(x^2)) ?
The derivative of (x^4-5x^3+8/(x^2)) is (2x^4-5x^3-16)/(x^3)
What is the first derivative of (x^4-5x^3+8/(x^2)) ?
The first derivative of (x^4-5x^3+8/(x^2)) is (2x^4-5x^3-16)/(x^3)