What is the derivative of f(x)=arccos(e^{2x}) ?

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

What is the first derivative of f(x)=arccos(e^{2x}) ?

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

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derivative of

f (x) = arccos (e^{2 x})

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (e^{2 x}) = e^{2 x} \cdot 2

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

Simplify

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

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

x^{3} - 8 x^{2} + 12 x, - x^{3} + 8 x^{2} - 12 x

y = \frac{7}{( 1 - x ^{2} ) ^{3}}

\frac{\partial}{\partial x} (\frac{1}{x ^{4}})

e^{x} (y - 1) d x + 2 (e^{x} + 4) d y = 0

x, 4

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