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

The derivative of f(x)=((x^3+9))/(x^3) is -(27)/(x^4)

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

The first derivative of f(x)=((x^3+9))/(x^3) is -(27)/(x^4)

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derivative of

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

- \frac{27}{x ^{4}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{3 x ^{2} x ^{3} - 3 x ^{2} ( x ^{3} + 9 )}{( x ^{3} ) ^{2}} : - \frac{27}{x ^{4}}

= - \frac{27}{x ^{4}}

x \to 0 lim (tan (3 x))

\int \frac{x - 2}{x} d x

d er i v a t i v e y = - 2 e^{- x^{2}} x

\int \frac{3 x}{1 + x ^{3}} d x

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

What is the derivative of f(x)=((x^3+9))/(x^3) ?
The derivative of f(x)=((x^3+9))/(x^3) is -(27)/(x^4)
What is the first derivative of f(x)=((x^3+9))/(x^3) ?
The first derivative of f(x)=((x^3+9))/(x^3) is -(27)/(x^4)