What is the derivative of sqrt(x^2+y^2-9) ?

The derivative of sqrt(x^2+y^2-9) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-9)}

What is the first derivative of sqrt(x^2+y^2-9) ?

The first derivative of sqrt(x^2+y^2-9) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-9)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (x^{2} + y^{2} - 9)

\frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9}

Solution steps

\frac{d}{d x} (x^{2} + y^{2} - 9)

Apply the chain rule:

\frac{1}{2 x ^{2} + y ^{2} - 9} \frac{d}{d x} (x^{2} + y^{2} - 9)

= \frac{1}{2 x ^{2} + y ^{2} - 9} \frac{d}{d x} (x^{2} + y^{2} - 9)

\frac{d}{d x} (x^{2} + y^{2} - 9) = 2 x + 2 y \frac{d y}{d x}

= \frac{1}{2 x ^{2} + y ^{2} - 9} (2 x + 2 y \frac{d y}{d x})

Simplify

\frac{1}{2 x ^{2} + y ^{2} - 9} (2 x + 2 y \frac{d y}{d x}) : \frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9}

= \frac{x + y \frac{d y}{d x}}{x ^{2} + y ^{2} - 9}

\frac{d}{d x} (sin^{2} (x) - cos (2 x))

x \to 4 + lim (5)

\int sin (ln (14 x)) d x

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

\int_{0}^{30} 4 x (30 - x) d x

What is the derivative of sqrt(x^2+y^2-9) ?
The derivative of sqrt(x^2+y^2-9) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-9)}
What is the first derivative of sqrt(x^2+y^2-9) ?
The first derivative of sqrt(x^2+y^2-9) is (x+y\frac{dy)/(dx)}{sqrt(x^2+y^2-9)}