What is the derivative of 4arccos(sqrt(1-x^2)) ?

The derivative of 4arccos(sqrt(1-x^2)) is 4/(sqrt(1-x^2))

What is the first derivative of 4arccos(sqrt(1-x^2)) ?

The first derivative of 4arccos(sqrt(1-x^2)) is 4/(sqrt(1-x^2))

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

\frac{4}{1 - x ^{2}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

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

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

Simplify

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

= \frac{4}{1 - x ^{2}}

\frac{\partial}{\partial y} (- 2 y^{2})

\int_{- 2}^{3} \frac{39}{x ^{4}} d x

f (θ) = cos (θ^{2})

\frac{\partial}{\partial x} (ln (y^{2}))

What is the derivative of 4arccos(sqrt(1-x^2)) ?
The derivative of 4arccos(sqrt(1-x^2)) is 4/(sqrt(1-x^2))
What is the first derivative of 4arccos(sqrt(1-x^2)) ?
The first derivative of 4arccos(sqrt(1-x^2)) is 4/(sqrt(1-x^2))