What is the derivative of f(x)=((4x^5+8))/(x^3) ?

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

What is the first derivative of f(x)=((4x^5+8))/(x^3) ?

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

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derivative of

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

\frac{8 ( x ^{5} - 3 )}{x ^{4}}

Solution steps

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

\frac{d}{d x} (4 x^{5} + 8) = 20 x^{4}

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

= \frac{20 x ^{4} x ^{3} - 3 x ^{2} ( 4 x ^{5} + 8 )}{( x ^{3} ) ^{2}}

Simplify

\frac{20 x ^{4} x ^{3} - 3 x ^{2} ( 4 x ^{5} + 8 )}{( x ^{3} ) ^{2}} : \frac{8 ( x ^{5} - 3 )}{x ^{4}}

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

f (x) = 9 x^{\frac{1}{3}}

y^{^{'}} 1 - 4 x^{2} = x

y^{4} y^{^{'}} = - 3 x^{2} y^{5} + x^{2}

f (t) = (8 t - 7)^{4} + (8 t + 7)^{4}

\frac{\partial}{\partial x} (\frac{( 7 x + 7 y )}{z})

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