What is the derivative of sqrt(1/4 x^2-2x-y) ?

The derivative of sqrt(1/4 x^2-2x-y) is (x-4-2\frac{dy)/(dx)}{2sqrt(x^2-8x-4y)}

What is the first derivative of sqrt(1/4 x^2-2x-y) ?

The first derivative of sqrt(1/4 x^2-2x-y) is (x-4-2\frac{dy)/(dx)}{2sqrt(x^2-8x-4y)}

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\frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

\frac{x - 4 - 2 \frac{d y}{d x}}{2 x ^{2} - 8 x - 4 y}

Solution steps

\frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

Apply the chain rule:

\frac{1}{2 \frac{1}{4} x ^{2} - 2 x - y} \frac{d}{d x} (\frac{1}{4} x^{2} - 2 x - y)

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

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

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

Simplify

\frac{1}{2 \frac{1}{4} x ^{2} - 2 x - y} (\frac{x}{2} - 2 - \frac{d y}{d x}) : \frac{x - 4 - 2 \frac{d y}{d x}}{2 x ^{2} - 8 x - 4 y}

= \frac{x - 4 - 2 \frac{d y}{d x}}{2 x ^{2} - 8 x - 4 y}

\frac{\partial}{\partial y} (ln (x y))

\int x^{2} 3 - 2 x d x

\int \frac{1}{( a ^{2} + v ^{2} )} d v

y^{^{'}} + tan (x) y - sin (x) = 0

\int \frac{1}{x} + 2 x d x

What is the derivative of sqrt(1/4 x^2-2x-y) ?
The derivative of sqrt(1/4 x^2-2x-y) is (x-4-2\frac{dy)/(dx)}{2sqrt(x^2-8x-4y)}
What is the first derivative of sqrt(1/4 x^2-2x-y) ?
The first derivative of sqrt(1/4 x^2-2x-y) is (x-4-2\frac{dy)/(dx)}{2sqrt(x^2-8x-4y)}