What is (\partial)/(\partial x)(cos(x^2+y^2-z)) ?

The answer to (\partial)/(\partial x)(cos(x^2+y^2-z)) is -sin(x^2+y^2-z)*2x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\partial)/(\partial x)(cos(x^2+y^2-z))

\frac{\partial}{\partial x} (cos (x^{2} + y^{2} - z))

- sin (x^{2} + y^{2} - z) \cdot 2 x

Solution steps

\frac{\partial}{\partial x} (cos (x^{2} + y^{2} - z))

Treat

y, z

as constants

Apply the chain rule:

- sin (x^{2} + y^{2} - z) \frac{\partial}{\partial x} (x^{2} + y^{2} - z)

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

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

= - sin (x^{2} + y^{2} - z) \cdot 2 x

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

\int - y^{2} e^{y} d y

\int (\frac{1}{4})^{x} d x

x \to - 3 lim (\frac{1 - 6 x}{3 + x})

\int_{0}^{1} \frac{x ^{2}}{x ^{3} + 1} d x

What is (\partial)/(\partial x)(cos(x^2+y^2-z)) ?
The answer to (\partial)/(\partial x)(cos(x^2+y^2-z)) is -sin(x^2+y^2-z)*2x