What is (\partial)/(\partial x)(2x^2+3y^2-z^2) ?

The answer to (\partial)/(\partial x)(2x^2+3y^2-z^2) is 4x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(2x^2+3y^2-z^2)

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} - z^{2})

4 x

Solution steps

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2} - z^{2})

Treat

y, z

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (2 x^{2}) + \frac{\partial}{\partial x} (3 y^{2}) - \frac{\partial}{\partial x} (z^{2})

\frac{\partial}{\partial x} (2 x^{2}) = 4 x

\frac{\partial}{\partial x} (3 y^{2}) = 0

\frac{\partial}{\partial x} (z^{2}) = 0

= 4 x + 0 - 0

Simplify

= 4 x

y = 4 + x x, (1, 5)

x \to 2 lim (x^{4} - 8)

y^{^{''}} + y^{^{'}} - 6 y = 12 x - 50 sin (x)

\frac{d ^{2}}{d x ^{2}} (cos (7 x))

y^{^{'}} = 3 x (y + 4)^{2}

What is (\partial)/(\partial x)(2x^2+3y^2-z^2) ?
The answer to (\partial)/(\partial x)(2x^2+3y^2-z^2) is 4x