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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int \frac{1}{1 + 2 cos ( x )} d x

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

y = 4 2 x, y = 2 x^{2}, y = - 4 x + 6

\int - 10 cos (- 2 x + 7) d x

f (x) = sin (sin (x))

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