What is the derivative of (3x^5-7x^2-4/(x^2)) ?

The derivative of (3x^5-7x^2-4/(x^2)) is (9x^5+8)/(x^3)

What is the first derivative of (3x^5-7x^2-4/(x^2)) ?

The first derivative of (3x^5-7x^2-4/(x^2)) is (9x^5+8)/(x^3)

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\frac{d}{d x} (\frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}})

\frac{9 x ^{5} + 8}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 x ^{5} - 7 x ^{2} - 4 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 3 x ^{5} - 7 x ^{2} - 4 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (3 x^{5} - 7 x^{2} - 4) = 15 x^{4} - 14 x

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

= \frac{( 15 x ^{4} - 14 x ) x ^{2} - 2 x ( 3 x ^{5} - 7 x ^{2} - 4 )}{( x ^{2} ) ^{2}}

Simplify

\frac{( 15 x ^{4} - 14 x ) x ^{2} - 2 x ( 3 x ^{5} - 7 x ^{2} - 4 )}{( x ^{2} ) ^{2}} : \frac{9 x ^{5} + 8}{x ^{3}}

= \frac{9 x ^{5} + 8}{x ^{3}}

sin (\frac{1}{x ^{2} + 1})

\frac{sin ( x ) - x}{x ^{2}}

(- 2 x y 2 - 1 y) d x + (- 2 x 2 y + 1 x) d y = 0

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

\int \frac{- 2}{1 + x ^{2}} d x

What is the derivative of (3x^5-7x^2-4/(x^2)) ?
The derivative of (3x^5-7x^2-4/(x^2)) is (9x^5+8)/(x^3)
What is the first derivative of (3x^5-7x^2-4/(x^2)) ?
The first derivative of (3x^5-7x^2-4/(x^2)) is (9x^5+8)/(x^3)