What is the derivative of f(x)=(x^3+9x^2+x-8)/x ?

The derivative of f(x)=(x^3+9x^2+x-8)/x is (2x^3+9x^2+8)/(x^2)

What is the first derivative of f(x)=(x^3+9x^2+x-8)/x ?

The first derivative of f(x)=(x^3+9x^2+x-8)/x is (2x^3+9x^2+8)/(x^2)

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derivative of

f (x) = \frac{x ^{3} + 9 x ^{2} + x - 8}{x}

\frac{2 x ^{3} + 9 x ^{2} + 8}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 9 x ^{2} + x - 8}{x})

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 ^{2} + x - 8 ) x - \frac{d x}{d x} ( x ^{3} + 9 x ^{2} + x - 8 )}{x ^{2}}

\frac{d}{d x} (x^{3} + 9 x^{2} + x - 8) = 3 x^{2} + 18 x + 1

\frac{d x}{d x} = 1

= \frac{( 3 x ^{2} + 18 x + 1 ) x - 1 \cdot ( x ^{3} + 9 x ^{2} + x - 8 )}{x ^{2}}

Simplify

\frac{( 3 x ^{2} + 18 x + 1 ) x - 1 \cdot ( x ^{3} + 9 x ^{2} + x - 8 )}{x ^{2}} : \frac{2 x ^{3} + 9 x ^{2} + 8}{x ^{2}}

= \frac{2 x ^{3} + 9 x ^{2} + 8}{x ^{2}}

\frac{\partial}{\partial x} (4 a x)

\frac{d}{d x} ((x^{2} + 6)^{9})

\frac{d}{d x} (cosh (4 x))

\int \frac{2 x ^{2} + 7 x - 8}{x - 7} d x

\frac{\partial}{\partial y} (e^{2 x + 3 y})

What is the derivative of f(x)=(x^3+9x^2+x-8)/x ?
The derivative of f(x)=(x^3+9x^2+x-8)/x is (2x^3+9x^2+8)/(x^2)
What is the first derivative of f(x)=(x^3+9x^2+x-8)/x ?
The first derivative of f(x)=(x^3+9x^2+x-8)/x is (2x^3+9x^2+8)/(x^2)