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

The derivative of (x^2-3x/(sqrt(x+1))) is (3x^2+x-6)/(2(x+1)sqrt(x+1))

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

The first derivative of (x^2-3x/(sqrt(x+1))) is (3x^2+x-6)/(2(x+1)sqrt(x+1))

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

\frac{3 x ^{2} + x - 6}{2 ( x + 1 ) x + 1}

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) = \frac{1}{2 x + 1}

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

Simplify

\frac{( 2 x - 3 ) x + 1 - \frac{1}{2 x + 1} ( x ^{2} - 3 x )}{( x + 1 ) ^{2}} : \frac{3 x ^{2} + x - 6}{2 ( x + 1 ) x + 1}

= \frac{3 x ^{2} + x - 6}{2 ( x + 1 ) x + 1}

\int 9 x e^{17 x} d x

\int cot^{3} (2 x) d x

(x^{2} + 4)^{7}

x \to a lim (2 - x)

y^{^{''}} - 6 y^{^{'}} + 9 y = 24 - 9 t - (12 t e^{3 t} + 4 e^{3 t})

What is the derivative of (x^2-3x/(sqrt(x+1))) ?
The derivative of (x^2-3x/(sqrt(x+1))) is (3x^2+x-6)/(2(x+1)sqrt(x+1))
What is the first derivative of (x^2-3x/(sqrt(x+1))) ?
The first derivative of (x^2-3x/(sqrt(x+1))) is (3x^2+x-6)/(2(x+1)sqrt(x+1))