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

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

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

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

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derivative of

y = arccos (4 x)

- \frac{4}{1 - 16 x ^{2}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (4 x) = 4

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

Simplify

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

= - \frac{4}{1 - 16 x ^{2}}

x \to \infty lim (x + 1)

(e^{t} cos (t) + e^{t} sin (t) + e^{t})^{^{'}}

cos^{2} (x)

\int 5^{x} ln (5) d x

\int_{- π}^{π} cos^{2} (3 x) d x

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