What is the derivative of y=(3x^4+9)/(x^2) ?

The derivative of y=(3x^4+9)/(x^2) is (6(x^4-3))/(x^3)

What is the first derivative of y=(3x^4+9)/(x^2) ?

The first derivative of y=(3x^4+9)/(x^2) is (6(x^4-3))/(x^3)

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derivative of

y = \frac{3 x ^{4} + 9}{x ^{2}}

\frac{6 ( x ^{4} - 3 )}{x ^{3}}

Solution steps

\frac{d}{d x} (\frac{3 x ^{4} + 9}{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 ^{4} + 9 ) x ^{2} - \frac{d}{d x} ( x ^{2} ) ( 3 x ^{4} + 9 )}{( x ^{2} ) ^{2}}

\frac{d}{d x} (3 x^{4} + 9) = 12 x^{3}

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

= \frac{12 x ^{3} x ^{2} - 2 x ( 3 x ^{4} + 9 )}{( x ^{2} ) ^{2}}

Simplify

\frac{12 x ^{3} x ^{2} - 2 x ( 3 x ^{4} + 9 )}{( x ^{2} ) ^{2}} : \frac{6 ( x ^{4} - 3 )}{x ^{3}}

= \frac{6 ( x ^{4} - 3 )}{x ^{3}}

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

\int (ln (x + 1))^{2} d x

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

\frac{d}{d x} (e^{12 x - x^{6}})

\int cot^{4} (5 x) d x

What is the derivative of y=(3x^4+9)/(x^2) ?
The derivative of y=(3x^4+9)/(x^2) is (6(x^4-3))/(x^3)
What is the first derivative of y=(3x^4+9)/(x^2) ?
The first derivative of y=(3x^4+9)/(x^2) is (6(x^4-3))/(x^3)