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

The derivative of (x^2+1/(x-9)) is (x^2-18x-1)/((x-9)^2)

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

The first derivative of (x^2+1/(x-9)) is (x^2-18x-1)/((x-9)^2)

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

\frac{x ^{2} - 18 x - 1}{( x - 9 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (x - 9) = 1

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

Simplify

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

= \frac{x ^{2} - 18 x - 1}{( x - 9 ) ^{2}}

\frac{\partial}{\partial x} (2 (x + y))

\int (1 + y) d y

\int \frac{1}{x ^{3}} e^{\frac{1}{x}} d x

\int v^{- 2} d v

\frac{d}{d x} (- a sin (a x))

What is the derivative of (x^2+1/(x-9)) ?
The derivative of (x^2+1/(x-9)) is (x^2-18x-1)/((x-9)^2)
What is the first derivative of (x^2+1/(x-9)) ?
The first derivative of (x^2+1/(x-9)) is (x^2-18x-1)/((x-9)^2)