What is implicit (dy)/(dx),x^2+y^2=0 ?

The answer to implicit (dy)/(dx),x^2+y^2=0 is (dy)/(dx)=-x/y

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

implicit (dy)/(dx),x^2+y^2=0

implicit derivative

\frac{d y}{d x}, x^{2} + y^{2} = 0

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

Solution steps

x^{2} + y^{2} = 0

Treat

y

y (x)

Differentiate both sides:

2 x + 2 y \frac{d y}{d x} = 0

Isolate

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

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

R (t) = (t + e^{t}) (3 - t)

\frac{\partial}{\partial x} (\frac{x}{x ^{3} - y ^{5}})

n = 0 \sum \infty \frac{7}{n ( n + 3 )}

x = - y^{2} + 6, y = \frac{x}{2} - \frac{3}{2}

y^{^{'}} + 8 y = x

What is implicit (dy)/(dx),x^2+y^2=0 ?
The answer to implicit (dy)/(dx),x^2+y^2=0 is (dy)/(dx)=-x/y