What is the derivative of 1-x/2-ln(2+9x-x^2) ?

The derivative of 1-x/2-ln(2+9x-x^2) is -(-2x+9)/(-x^2+9x+2)-1/2

What is the first derivative of 1-x/2-ln(2+9x-x^2) ?

The first derivative of 1-x/2-ln(2+9x-x^2) is -(-2x+9)/(-x^2+9x+2)-1/2

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\frac{d}{d x} (1 - \frac{x}{2} - ln (2 + 9 x - x^{2}))

- \frac{- 2 x + 9}{- x ^{2} + 9 x + 2} - \frac{1}{2}

Solution steps

\frac{d}{d x} (1 - \frac{x}{2} - ln (2 + 9 x - x^{2}))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (1) - \frac{d}{d x} (\frac{x}{2}) - \frac{d}{d x} (ln (2 + 9 x - x^{2}))

\frac{d}{d x} (1) = 0

\frac{d}{d x} (\frac{x}{2}) = \frac{1}{2}

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

= 0 - \frac{1}{2} - \frac{- 2 x + 9}{2 + 9 x - x ^{2}}

Simplify

= - \frac{- 2 x + 9}{- x ^{2} + 9 x + 2} - \frac{1}{2}

\int e^{x} sin (x) sin (3 x) d x

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

\frac{d}{d x} (\frac{cos ( x ) - 1}{x ^{2}})

\int_{2}^{4} (2 + u)^{2} d u

\frac{d}{d x} (- ln (x) cos (x))

What is the derivative of 1-x/2-ln(2+9x-x^2) ?
The derivative of 1-x/2-ln(2+9x-x^2) is -(-2x+9)/(-x^2+9x+2)-1/2
What is the first derivative of 1-x/2-ln(2+9x-x^2) ?
The first derivative of 1-x/2-ln(2+9x-x^2) is -(-2x+9)/(-x^2+9x+2)-1/2