What is the derivative of (6x^4/(x^2-3)) ?

The derivative of (6x^4/(x^2-3)) is (6(2x^5-12x^3))/((x^2-3)^2)

What is the first derivative of (6x^4/(x^2-3)) ?

The first derivative of (6x^4/(x^2-3)) is (6(2x^5-12x^3))/((x^2-3)^2)

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\frac{d}{d x} (\frac{6 x ^{4}}{x ^{2} - 3})

\frac{6 ( 2 x ^{5} - 12 x ^{3} )}{( x ^{2} - 3 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x ^{4}}{x ^{2} - 3})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d}{d x} ( x ^{4} ) ( x ^{2} - 3 ) - \frac{d}{d x} ( x ^{2} - 3 ) x ^{4}}{( x ^{2} - 3 ) ^{2}}

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

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

= 6 \cdot \frac{4 x ^{3} ( x ^{2} - 3 ) - 2 x x ^{4}}{( x ^{2} - 3 ) ^{2}}

Simplify

6 \cdot \frac{4 x ^{3} ( x ^{2} - 3 ) - 2 x x ^{4}}{( x ^{2} - 3 ) ^{2}} : \frac{6 ( 2 x ^{5} - 12 x ^{3} )}{( x ^{2} - 3 ) ^{2}}

= \frac{6 ( 2 x ^{5} - 12 x ^{3} )}{( x ^{2} - 3 ) ^{2}}

f (x) = cos (2 \cdot x) \cdot sin (5 \cdot x)

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

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

n = 1 \sum \infty \frac{n}{( 2 n ) !}

f (x) = x^{2} - 8 x - 8

What is the derivative of (6x^4/(x^2-3)) ?
The derivative of (6x^4/(x^2-3)) is (6(2x^5-12x^3))/((x^2-3)^2)
What is the first derivative of (6x^4/(x^2-3)) ?
The first derivative of (6x^4/(x^2-3)) is (6(2x^5-12x^3))/((x^2-3)^2)