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

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

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

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

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

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

Solution steps

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

\frac{d}{d x} (3 x^{4} + 7) = 12 x^{3}

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

= \frac{12 x ^{3} x ^{2} - 2 x ( 3 x ^{4} + 7 )}{( x ^{2} ) ^{2}}

Simplify

\frac{12 x ^{3} x ^{2} - 2 x ( 3 x ^{4} + 7 )}{( x ^{2} ) ^{2}} : \frac{2 ( 3 x ^{4} - 7 )}{x ^{3}}

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

\int 2 x (x^{2} + 6)^{6} d x

R \to 1 + lim (\frac{1}{R - 1})

a re a \frac{2}{x}, 8 x, \frac{1}{8 x}

\int (3 x - 3) e^{3 x^{2} - 6 x} d x

d er i v a t i v e y = 1 7^{x}

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