What is the derivative of y=arccos(4x^2) ?

The derivative of y=arccos(4x^2) is -(8x)/(sqrt(1-16x^4))

What is the first derivative of y=arccos(4x^2) ?

The first derivative of y=arccos(4x^2) is -(8x)/(sqrt(1-16x^4))

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derivative of

y = arccos (4 x^{2})

- \frac{8 x}{1 - 16 x ^{4}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (4 x^{2}) = 8 x

= - \frac{1}{1 - ( 4 x ^{2} ) ^{2}} \cdot 8 x

Simplify

- \frac{1}{1 - ( 4 x ^{2} ) ^{2}} \cdot 8 x : - \frac{8 x}{1 - 16 x ^{4}}

= - \frac{8 x}{1 - 16 x ^{4}}

y = x^{2}, y = x + 2

\int (9 - x^{2})^{\frac{3}{2}} d x

\int \frac{1}{36 x ^{2} + 1} d x

(tan (θ))^{^{'}}

x = y^{3}, x = y^{2}

What is the derivative of y=arccos(4x^2) ?
The derivative of y=arccos(4x^2) is -(8x)/(sqrt(1-16x^4))
What is the first derivative of y=arccos(4x^2) ?
The first derivative of y=arccos(4x^2) is -(8x)/(sqrt(1-16x^4))