What is the derivative of (arccos(x)/(e^x)) ?

The derivative of (arccos(x)/(e^x)) is -(1+arccos(x)sqrt(-x^2+1))/(e^xsqrt(1-x^2))

What is the first derivative of (arccos(x)/(e^x)) ?

The first derivative of (arccos(x)/(e^x)) is -(1+arccos(x)sqrt(-x^2+1))/(e^xsqrt(1-x^2))

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

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

Solution steps

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

\frac{( - \frac{1}{1 - x ^{2}} ) e ^{x} - e ^{x} arccos ( x )}{( e ^{x} ) ^{2}} : - \frac{1 + arccos ( x ) - x ^{2} + 1}{e ^{x} 1 - x ^{2}}

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

- \frac{d y}{d x} = 2 y - x

\int \frac{e ^{x^{2}}}{2 x} \cdot 5 x^{4} d x

\frac{d}{d t} ((sin (ln (t)))^{2})

\int_{1}^{10} \frac{29}{x ^{3}} d x

\int (x^{2} + ln (2 x) + e^{- 2 x}) d x

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