What is (\partial)/(\partial y)(tan(x^2+y^2)) ?

The answer to (\partial)/(\partial y)(tan(x^2+y^2)) is sec^2(x^2+y^2)*2y

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(tan(x^2+y^2))

\frac{\partial}{\partial y} (tan (x^{2} + y^{2}))

sec^{2} (x^{2} + y^{2}) \cdot 2 y

Solution steps

\frac{\partial}{\partial y} (tan (x^{2} + y^{2}))

Treat

x

as a constant

Apply the chain rule:

sec^{2} (x^{2} + y^{2}) \frac{\partial}{\partial y} (x^{2} + y^{2})

= sec^{2} (x^{2} + y^{2}) \frac{\partial}{\partial y} (x^{2} + y^{2})

\frac{\partial}{\partial y} (x^{2} + y^{2}) = 2 y

= sec^{2} (x^{2} + y^{2}) \cdot 2 y

f (x) = \frac{1 - sin ( x )}{1 + sin ( x )}

\int \frac{1}{x + 4} d x

f (x) = x - a

\frac{d}{d x} (x ln (x) - x)

\frac{\partial}{\partial y} (x cos (y - x))

What is (\partial)/(\partial y)(tan(x^2+y^2)) ?
The answer to (\partial)/(\partial y)(tan(x^2+y^2)) is sec^2(x^2+y^2)*2y