What is the derivative of (x^2+x+4/(x+1)) ?

The derivative of (x^2+x+4/(x+1)) is (x^2+2x-3)/((x+1)^2)

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

The first derivative of (x^2+x+4/(x+1)) is (x^2+2x-3)/((x+1)^2)

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

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

Solution steps

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

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

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

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

Simplify

\frac{( 2 x + 1 ) ( x + 1 ) - 1 \cdot ( x ^{2} + x + 4 )}{( x + 1 ) ^{2}} : \frac{x ^{2} + 2 x - 3}{( x + 1 ) ^{2}}

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

\int \frac{x ^{5} - 3 x}{x ^{4}} d x

\int 4 y + 2 x - 5 d x

y = x^{2} + 1, x = 0, x = 6

\int - 4 sin (2 x) + 3 d x

\frac{d}{d x} (\frac{6 x - 1}{4 x + 5})

What is the derivative of (x^2+x+4/(x+1)) ?
The derivative of (x^2+x+4/(x+1)) is (x^2+2x-3)/((x+1)^2)
What is the first derivative of (x^2+x+4/(x+1)) ?
The first derivative of (x^2+x+4/(x+1)) is (x^2+2x-3)/((x+1)^2)