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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{2 x}{x ^{2} + y}

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= \frac{2 x}{x ^{2} + y} - 0

Simplify

= \frac{2 x}{x ^{2} + y}

x \to 3 lim (x^{2} + 9)

2 sin^{2} (x)

f (y) = \frac{1}{2 y}

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

\int x^{4} + x + 9 d x

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