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

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

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

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

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derivative of

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

\frac{6 ( x ^{4} + 9 )}{x ^{4}}

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

= \frac{6 ( x ^{4} + 9 )}{x ^{4}}

f (x) = \frac{x}{x + 6}

y = 6 x, y = 2 x - 6, y = 0

\frac{1}{x}, 1

f (x) = x^{2} - x, at x = 2

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

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