What is the derivative of f(x)=(1300+8x-0.0004x^2)/x ?

The derivative of f(x)=(1300+8x-0.0004x^2)/x is (-0.0004x^2-1300)/(x^2)

What is the first derivative of f(x)=(1300+8x-0.0004x^2)/x ?

The first derivative of f(x)=(1300+8x-0.0004x^2)/x is (-0.0004x^2-1300)/(x^2)

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derivative of

f (x) = \frac{1300 + 8 x - 0.0004 x ^{2}}{x}

\frac{- 0.0004 x ^{2} - 1300}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{1300 + 8 x - 0.0004 x ^{2}}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 1300 + 8 x - 0.0004 x ^{2} ) x - \frac{d x}{d x} ( 1300 + 8 x - 0.0004 x ^{2} )}{x ^{2}}

\frac{d}{d x} (1300 + 8 x - 0.0004 x^{2}) = - 0.0008 x + 8

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

= \frac{( - 0.0008 x + 8 ) x - 1 \cdot ( 1300 + 8 x - 0.0004 x ^{2} )}{x ^{2}}

Simplify

\frac{( - 0.0008 x + 8 ) x - 1 \cdot ( 1300 + 8 x - 0.0004 x ^{2} )}{x ^{2}} : \frac{- 0.0004 x ^{2} - 1300}{x ^{2}}

= \frac{- 0.0004 x ^{2} - 1300}{x ^{2}}

\frac{s + 1}{s ^{2} + 1}

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

f^{^{''}} (x) = 2 x

\int_{0}^{l n (2)} \frac{4}{e ^{x} + 3} d x

\frac{d y}{d x} x^{2} + y x = 1, y (6) = 9

What is the derivative of f(x)=(1300+8x-0.0004x^2)/x ?
The derivative of f(x)=(1300+8x-0.0004x^2)/x is (-0.0004x^2-1300)/(x^2)
What is the first derivative of f(x)=(1300+8x-0.0004x^2)/x ?
The first derivative of f(x)=(1300+8x-0.0004x^2)/x is (-0.0004x^2-1300)/(x^2)