What is the derivative of arccos(x+y) ?

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

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

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

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\frac{d}{d x} (arccos (x + y))

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

Solution steps

\frac{d}{d x} (arccos (x + y))

Apply the chain rule:

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

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

\frac{d}{d x} (x + y) = 1 + \frac{d y}{d x}

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

Simplify

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

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

t an g e n t f (x) = 3 x^{2}, at x = (6)

x \to 1 + lim (\frac{1}{x ^{3} - 1})

d er i v a t i v e f (x) = tan^{3} (x)

\int sin^{0} (x) d x

\int_{0}^{8} (8 y - y^{2}) d y

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