What is the derivative of (2-1/x /(x-3)) ?

The derivative of (2-1/x /(x-3)) is (-2x^2+2x-3)/(x^2(x-3)^2)

What is the first derivative of (2-1/x /(x-3)) ?

The first derivative of (2-1/x /(x-3)) is (-2x^2+2x-3)/(x^2(x-3)^2)

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\frac{d}{d x} (\frac{2 - \frac{1}{x}}{x - 3})

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

Solution steps

\frac{d}{d x} (\frac{2 - \frac{1}{x}}{x - 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 2 - \frac{1}{x} ) ( x - 3 ) - \frac{d}{d x} ( x - 3 ) ( 2 - \frac{1}{x} )}{( x - 3 ) ^{2}}

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

\frac{d}{d x} (x - 3) = 1

= \frac{\frac{1}{x ^{2}} ( x - 3 ) - 1 \cdot ( 2 - \frac{1}{x} )}{( x - 3 ) ^{2}}

Simplify

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{2 x} (3 - 5 x)

\int \frac{x + 4}{2 x ^{2} + x - 1} d x

f (x) = x + 15, at x = 1

(3 x^{2}) d x + (2 y - 4 x^{3} y^{- 1}) d y = 0

\int x sin^{2} (5 x) d x

What is the derivative of (2-1/x /(x-3)) ?
The derivative of (2-1/x /(x-3)) is (-2x^2+2x-3)/(x^2(x-3)^2)
What is the first derivative of (2-1/x /(x-3)) ?
The first derivative of (2-1/x /(x-3)) is (-2x^2+2x-3)/(x^2(x-3)^2)