What is the derivative of y=(2x^5+6)/(x^3) ?

The derivative of y=(2x^5+6)/(x^3) is (2(2x^5-9))/(x^4)

What is the first derivative of y=(2x^5+6)/(x^3) ?

The first derivative of y=(2x^5+6)/(x^3) is (2(2x^5-9))/(x^4)

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derivative of

y = \frac{2 x ^{5} + 6}{x ^{3}}

\frac{2 ( 2 x ^{5} - 9 )}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{2 x ^{5} + 6}{x ^{3}})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x^{5} + 6) = 10 x^{4}

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

= \frac{10 x ^{4} x ^{3} - 3 x ^{2} ( 2 x ^{5} + 6 )}{( x ^{3} ) ^{2}}

Simplify

\frac{10 x ^{4} x ^{3} - 3 x ^{2} ( 2 x ^{5} + 6 )}{( x ^{3} ) ^{2}} : \frac{2 ( 2 x ^{5} - 9 )}{x ^{4}}

= \frac{2 ( 2 x ^{5} - 9 )}{x ^{4}}

\int_{0}^{5} 5 x d x

y = x, (64, 8)

\frac{12 s}{12 s - 30 s ^{2}}

\frac{d y}{d x} = \frac{( e ^{y} )}{x y}

\frac{1 - 5 t}{4 + t}

What is the derivative of y=(2x^5+6)/(x^3) ?
The derivative of y=(2x^5+6)/(x^3) is (2(2x^5-9))/(x^4)
What is the first derivative of y=(2x^5+6)/(x^3) ?
The first derivative of y=(2x^5+6)/(x^3) is (2(2x^5-9))/(x^4)