What is the derivative of f(x)=(2x^2+6x+6)/(x+1) ?

The derivative of f(x)=(2x^2+6x+6)/(x+1) is (2x^2+4x)/((x+1)^2)

What is the first derivative of f(x)=(2x^2+6x+6)/(x+1) ?

The first derivative of f(x)=(2x^2+6x+6)/(x+1) is (2x^2+4x)/((x+1)^2)

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derivative of

f (x) = \frac{2 x ^{2} + 6 x + 6}{x + 1}

\frac{2 x ^{2} + 4 x}{( x + 1 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x ^{2} + 6 x + 6}{x + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 2 x ^{2} + 6 x + 6 ) ( x + 1 ) - \frac{d}{d x} ( x + 1 ) ( 2 x ^{2} + 6 x + 6 )}{( x + 1 ) ^{2}}

\frac{d}{d x} (2 x^{2} + 6 x + 6) = 4 x + 6

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

= \frac{( 4 x + 6 ) ( x + 1 ) - 1 \cdot ( 2 x ^{2} + 6 x + 6 )}{( x + 1 ) ^{2}}

Simplify

\frac{( 4 x + 6 ) ( x + 1 ) - 1 \cdot ( 2 x ^{2} + 6 x + 6 )}{( x + 1 ) ^{2}} : \frac{2 x ^{2} + 4 x}{( x + 1 ) ^{2}}

= \frac{2 x ^{2} + 4 x}{( x + 1 ) ^{2}}

\frac{d y}{d x} = e^{4 y} sin (5 x)

\frac{\partial}{\partial x} (6 x + 4 y)

\int (x^{4})^{\frac{1}{2}} d x

h \to \frac{π}{2} lim (\frac{( 1 - cos ( 3 h ) )}{h})

What is the derivative of f(x)=(2x^2+6x+6)/(x+1) ?
The derivative of f(x)=(2x^2+6x+6)/(x+1) is (2x^2+4x)/((x+1)^2)
What is the first derivative of f(x)=(2x^2+6x+6)/(x+1) ?
The first derivative of f(x)=(2x^2+6x+6)/(x+1) is (2x^2+4x)/((x+1)^2)