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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

y^{^{''}} - y^{^{'}} + y = 0

\frac{d y}{d x}, x^{8} - y = 0

\int x sin (4 x) d x

\frac{\partial}{\partial x} ((\frac{y}{x}) \cdot ln (x))

x \to 0 lim (\frac{x}{x - 4})

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