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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x^{4} + 3 x^{2} - 1) = 8 x^{3} + 6 x

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

= \frac{( 8 x ^{3} + 6 x ) x ^{2} - 2 x ( 2 x ^{4} + 3 x ^{2} - 1 )}{( x ^{2} ) ^{2}}

Simplify

\frac{( 8 x ^{3} + 6 x ) x ^{2} - 2 x ( 2 x ^{4} + 3 x ^{2} - 1 )}{( x ^{2} ) ^{2}} : \frac{2 ( 2 x ^{4} + 1 )}{x ^{3}}

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

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

\int 8 x^{5} e^{5 x^{6}} d x

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = 0

y = 9 ln (5 x), y = 11, [2, 4]

\int sin^{2} (x) d x

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