What is the derivative of 4e^{arccos(4x)} ?

The derivative of 4e^{arccos(4x)} is -(16e^{arccos(4x)})/(sqrt(1-16x^2))

What is the first derivative of 4e^{arccos(4x)} ?

The first derivative of 4e^{arccos(4x)} is -(16e^{arccos(4x)})/(sqrt(1-16x^2))

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derivative of

4 e^{a r c c o s (4 x)}

- \frac{16 e ^{a r c c o s (4 x)}}{1 - 16 x ^{2}}

Solution steps

\frac{d}{d x} (4 e^{a r c c o s (4 x)})

Take the constant out:

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

= 4 \frac{d}{d x} (e^{a r c c o s (4 x)})

Apply the chain rule:

e^{a r c c o s (4 x)} \frac{d}{d x} (arccos (4 x))

= e^{a r c c o s (4 x)} \frac{d}{d x} (arccos (4 x))

\frac{d}{d x} (arccos (4 x)) = - \frac{4}{1 - 16 x ^{2}}

= 4 e^{a r c c o s (4 x)} (- \frac{4}{1 - 16 x ^{2}})

Simplify

4 e^{a r c c o s (4 x)} (- \frac{4}{1 - 16 x ^{2}}) : - \frac{16 e ^{a r c c o s (4 x)}}{1 - 16 x ^{2}}

= - \frac{16 e ^{a r c c o s (4 x)}}{1 - 16 x ^{2}}

\int 3 x^{3} d x

f (x) = ln (x^{2} - 2 x + 1), at x = 2

y^{^{'''}} - y^{^{''}} + y^{^{'}} - y = 0

g (x) = 7 - x

x \to 4 lim (3)

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