What is the derivative of arccos(2xsqrt(1-x^2)) ?

The derivative of arccos(2xsqrt(1-x^2)) is -(2(-2x^2+1))/(sqrt(1-4x^2(1-x^2))\sqrt{1-x^2)}

What is the first derivative of arccos(2xsqrt(1-x^2)) ?

The first derivative of arccos(2xsqrt(1-x^2)) is -(2(-2x^2+1))/(sqrt(1-4x^2(1-x^2))\sqrt{1-x^2)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (e^{- 2 t})

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

(2.5) (3.7)

y = - \frac{2}{x ^{1.1}} + \frac{9}{x ^{- 2.8}}

\int \frac{1}{2 + 3 x} d x

What is the derivative of arccos(2xsqrt(1-x^2)) ?
The derivative of arccos(2xsqrt(1-x^2)) is -(2(-2x^2+1))/(sqrt(1-4x^2(1-x^2))\sqrt{1-x^2)}
What is the first derivative of arccos(2xsqrt(1-x^2)) ?
The first derivative of arccos(2xsqrt(1-x^2)) is -(2(-2x^2+1))/(sqrt(1-4x^2(1-x^2))\sqrt{1-x^2)}