What is the derivative of arccos(sqrt(y-2x)) ?

The derivative of arccos(sqrt(y-2x)) is -(\frac{dy)/(dx)-2}{2sqrt(1-y+2x)sqrt(y-2x)}

What is the first derivative of arccos(sqrt(y-2x)) ?

The first derivative of arccos(sqrt(y-2x)) is -(\frac{dy)/(dx)-2}{2sqrt(1-y+2x)sqrt(y-2x)}

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\frac{d}{d x} (arccos (y - 2 x))

- \frac{\frac{d y}{d x} - 2}{2 1 - y + 2 x y - 2 x}

Solution steps

\frac{d}{d x} (arccos (y - 2 x))

Apply the chain rule:

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

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

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

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

Simplify

- \frac{1}{1 - ( y - 2 x ) ^{2}} \cdot \frac{\frac{d y}{d x} - 2}{2 y - 2 x} : - \frac{\frac{d y}{d x} - 2}{2 1 - y + 2 x y - 2 x}

= - \frac{\frac{d y}{d x} - 2}{2 1 - y + 2 x y - 2 x}

\int sec^{2} (θ) tan^{7} (θ) d θ

x \to 2 lim (\frac{- 1}{( x + 2 )})

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

x \to 0 lim (16 - 7 x)

\int - cos (x) + C d x

What is the derivative of arccos(sqrt(y-2x)) ?
The derivative of arccos(sqrt(y-2x)) is -(\frac{dy)/(dx)-2}{2sqrt(1-y+2x)sqrt(y-2x)}
What is the first derivative of arccos(sqrt(y-2x)) ?
The first derivative of arccos(sqrt(y-2x)) is -(\frac{dy)/(dx)-2}{2sqrt(1-y+2x)sqrt(y-2x)}