What is the derivative of f(x)=(x^3+8x^2+x-7)/x ?

The derivative of f(x)=(x^3+8x^2+x-7)/x is (2x^3+8x^2+7)/(x^2)

What is the first derivative of f(x)=(x^3+8x^2+x-7)/x ?

The first derivative of f(x)=(x^3+8x^2+x-7)/x is (2x^3+8x^2+7)/(x^2)

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derivative of

f (x) = \frac{x ^{3} + 8 x ^{2} + x - 7}{x}

\frac{2 x ^{3} + 8 x ^{2} + 7}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{3} + 8 x ^{2} + x - 7}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( x ^{3} + 8 x ^{2} + x - 7 ) x - \frac{d x}{d x} ( x ^{3} + 8 x ^{2} + x - 7 )}{x ^{2}}

\frac{d}{d x} (x^{3} + 8 x^{2} + x - 7) = 3 x^{2} + 16 x + 1

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

= \frac{( 3 x ^{2} + 16 x + 1 ) x - 1 \cdot ( x ^{3} + 8 x ^{2} + x - 7 )}{x ^{2}}

Simplify

\frac{( 3 x ^{2} + 16 x + 1 ) x - 1 \cdot ( x ^{3} + 8 x ^{2} + x - 7 )}{x ^{2}} : \frac{2 x ^{3} + 8 x ^{2} + 7}{x ^{2}}

= \frac{2 x ^{3} + 8 x ^{2} + 7}{x ^{2}}

n = 1 \sum \infty \frac{3}{( n + 1 ) !}

x \to 0 lim (\frac{3}{x} - \frac{3}{x ^{2} + x})

4

y^{^{'}} = \frac{3 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2}}{x ^{5} y}

\frac{d}{d x} ((x + 1)^{\frac{1}{2}})

What is the derivative of f(x)=(x^3+8x^2+x-7)/x ?
The derivative of f(x)=(x^3+8x^2+x-7)/x is (2x^3+8x^2+7)/(x^2)
What is the first derivative of f(x)=(x^3+8x^2+x-7)/x ?
The first derivative of f(x)=(x^3+8x^2+x-7)/x is (2x^3+8x^2+7)/(x^2)