What is the derivative of y=((3x^5+9))/(x^3) ?

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

What is the first derivative of y=((3x^5+9))/(x^3) ?

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

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derivative of

y = \frac{( 3 x ^{5} + 9 )}{x ^{3}}

\frac{3 ( 2 x ^{5} - 9 )}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{3 x ^{5} + 9}{x ^{3}})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 x ^{5} + 9 ) x ^{3} - \frac{d}{d x} ( x ^{3} ) ( 3 x ^{5} + 9 )}{( x ^{3} ) ^{2}}

\frac{d}{d x} (3 x^{5} + 9) = 15 x^{4}

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

= \frac{15 x ^{4} x ^{3} - 3 x ^{2} ( 3 x ^{5} + 9 )}{( x ^{3} ) ^{2}}

Simplify

\frac{15 x ^{4} x ^{3} - 3 x ^{2} ( 3 x ^{5} + 9 )}{( x ^{3} ) ^{2}} : \frac{3 ( 2 x ^{5} - 9 )}{x ^{4}}

= \frac{3 ( 2 x ^{5} - 9 )}{x ^{4}}

\frac{d}{d s} (ln (s^{2} + 9))

4 t^{- \frac{1}{8}}

r (x) = \frac{3}{x ^{4} + 2}

f (x) = \frac{6}{x ^{3}}

\frac{d}{d x} (sin (e^{3 x}))

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