What is the derivative of sqrt(3x^2+y^2-1) ?

The derivative of sqrt(3x^2+y^2-1) is (3x+y\frac{dy)/(dx)}{sqrt(3x^2+y^2-1)}

What is the first derivative of sqrt(3x^2+y^2-1) ?

The first derivative of sqrt(3x^2+y^2-1) is (3x+y\frac{dy)/(dx)}{sqrt(3x^2+y^2-1)}

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Popular >

\frac{d}{d x} (3 x^{2} + y^{2} - 1)

\frac{3 x + y \frac{d y}{d x}}{3 x ^{2} + y ^{2} - 1}

Solution steps

\frac{d}{d x} (3 x^{2} + y^{2} - 1)

Apply the chain rule:

\frac{1}{2 3 x ^{2} + y ^{2} - 1} \frac{d}{d x} (3 x^{2} + y^{2} - 1)

= \frac{1}{2 3 x ^{2} + y ^{2} - 1} \frac{d}{d x} (3 x^{2} + y^{2} - 1)

\frac{d}{d x} (3 x^{2} + y^{2} - 1) = 6 x + 2 y \frac{d y}{d x}

= \frac{1}{2 3 x ^{2} + y ^{2} - 1} (6 x + 2 y \frac{d y}{d x})

Simplify

\frac{1}{2 3 x ^{2} + y ^{2} - 1} (6 x + 2 y \frac{d y}{d x}) : \frac{3 x + y \frac{d y}{d x}}{3 x ^{2} + y ^{2} - 1}

= \frac{3 x + y \frac{d y}{d x}}{3 x ^{2} + y ^{2} - 1}

\int_{0}^{1} (x - x^{4}) d x

y = 3 x, y = 7 x^{2}

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

\frac{d}{d x} (\frac{2 x}{5 - tan ( x )})

\frac{d}{d x} (\frac{2 x + 5}{3 x - 1})

What is the derivative of sqrt(3x^2+y^2-1) ?
The derivative of sqrt(3x^2+y^2-1) is (3x+y\frac{dy)/(dx)}{sqrt(3x^2+y^2-1)}
What is the first derivative of sqrt(3x^2+y^2-1) ?
The first derivative of sqrt(3x^2+y^2-1) is (3x+y\frac{dy)/(dx)}{sqrt(3x^2+y^2-1)}