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

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

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

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

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Popular >

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x y^{^{'}} - 2 y = 11 x^{2}

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

\frac{d}{d x} (- \frac{3}{2} x)

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

\frac{d}{d x} (r^{2} - 17 x^{2})

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