What is the derivative of (-x^2+9x-16/x) ?

The derivative of (-x^2+9x-16/x) is (-x^2+16)/(x^2)

What is the first derivative of (-x^2+9x-16/x) ?

The first derivative of (-x^2+9x-16/x) is (-x^2+16)/(x^2)

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\frac{d}{d x} (\frac{- x ^{2} + 9 x - 16}{x})

\frac{- x ^{2} + 16}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{- x ^{2} + 9 x - 16}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( - x ^{2} + 9 x - 16 ) x - \frac{d x}{d x} ( - x ^{2} + 9 x - 16 )}{x ^{2}}

\frac{d}{d x} (- x^{2} + 9 x - 16) = - 2 x + 9

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

= \frac{( - 2 x + 9 ) x - 1 \cdot ( - x ^{2} + 9 x - 16 )}{x ^{2}}

Simplify

\frac{( - 2 x + 9 ) x - 1 \cdot ( - x ^{2} + 9 x - 16 )}{x ^{2}} : \frac{- x ^{2} + 16}{x ^{2}}

= \frac{- x ^{2} + 16}{x ^{2}}

\int \frac{9 - w ^{2}}{w ^{2}} d w

\frac{d}{d y} (0)

\frac{d}{d x} (tan (x) - e^{- x})

\int_{0}^{1} e^{3 x + 1} d x

\frac{d}{d x} (\frac{( x ^{2} + 4 )}{x})

What is the derivative of (-x^2+9x-16/x) ?
The derivative of (-x^2+9x-16/x) is (-x^2+16)/(x^2)
What is the first derivative of (-x^2+9x-16/x) ?
The first derivative of (-x^2+9x-16/x) is (-x^2+16)/(x^2)