What is the derivative of (3x^5+8/(x^3)) ?

The derivative of (3x^5+8/(x^3)) is (6(x^5-4))/(x^4)

What is the first derivative of (3x^5+8/(x^3)) ?

The first derivative of (3x^5+8/(x^3)) is (6(x^5-4))/(x^4)

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\frac{d}{d x} (\frac{3 x ^{5} + 8}{x ^{3}})

\frac{6 ( x ^{5} - 4 )}{x ^{4}}

Solution steps

\frac{d}{d x} (\frac{3 x ^{5} + 8}{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} + 8 ) x ^{3} - \frac{d}{d x} ( x ^{3} ) ( 3 x ^{5} + 8 )}{( x ^{3} ) ^{2}}

\frac{d}{d x} (3 x^{5} + 8) = 15 x^{4}

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

= \frac{15 x ^{4} x ^{3} - 3 x ^{2} ( 3 x ^{5} + 8 )}{( x ^{3} ) ^{2}}

Simplify

\frac{15 x ^{4} x ^{3} - 3 x ^{2} ( 3 x ^{5} + 8 )}{( x ^{3} ) ^{2}} : \frac{6 ( x ^{5} - 4 )}{x ^{4}}

= \frac{6 ( x ^{5} - 4 )}{x ^{4}}

\int 2 x ln (2 x) d x

y^{^{''}} - 6 y^{^{'}} + 10 y = 0

d er i v a t i v e C (x) = \frac{4 x + 50}{x + 2}

\int (x + 3) (x^{2} + 1) d x

\int 4 arcsin (x) d x

What is the derivative of (3x^5+8/(x^3)) ?
The derivative of (3x^5+8/(x^3)) is (6(x^5-4))/(x^4)
What is the first derivative of (3x^5+8/(x^3)) ?
The first derivative of (3x^5+8/(x^3)) is (6(x^5-4))/(x^4)