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

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

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

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

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derivative of

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

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

Solution steps

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

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

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

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

Simplify

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

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

2 t x^{^{''}} - x^{^{'}} + \frac{1}{( x ^{^{'}} )} = 0

\int \frac{( x ^{3} )}{x ^{2} + 16} d x

\int \frac{4 + 6 u}{u} d u

\int (x^{2} + 4 x) cos (x) d x

\int 3 x e^{0.5 x^{2}} d x

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