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

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

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

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

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derivative of

y = \frac{x ^{2} - 2}{2 x + 1}

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

Solution steps

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

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

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

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

Expand

2 x (2 x + 1) - 2 (x^{2} - 2) : 2 x^{2} + 2 x + 4

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

\int_{0}^{2} x - x^{2} d x

\int sin^{5} (θ) cos (θ) d θ

x^{2} y^{2} = 16, (1, 4)

\frac{\partial}{\partial x} (5 - 3 x^{2} - y^{2})

\frac{d y}{d x} - x^{2} y = 0

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