What is (\partial)/(\partial x)(ln(1+x^2+e^y+z)) ?

The answer to (\partial)/(\partial x)(ln(1+x^2+e^y+z)) is (2x)/(1+x^2+e^y+z)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(ln(1+x^2+e^y+z))

\frac{\partial}{\partial x} (ln (1 + x^{2} + e^{y} + z))

\frac{2 x}{1 + x ^{2} + e ^{y} + z}

Solution steps

\frac{\partial}{\partial x} (ln (1 + x^{2} + e^{y} + z))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{1 + x ^{2} + e ^{y} + z} \frac{\partial}{\partial x} (1 + x^{2} + e^{y} + z)

= \frac{1}{1 + x ^{2} + e ^{y} + z} \frac{\partial}{\partial x} (1 + x^{2} + e^{y} + z)

\frac{\partial}{\partial x} (1 + x^{2} + e^{y} + z) = 2 x

= \frac{1}{1 + x ^{2} + e ^{y} + z} \cdot 2 x

Simplify

\frac{1}{1 + x ^{2} + e ^{y} + z} \cdot 2 x : \frac{2 x}{1 + x ^{2} + e ^{y} + z}

= \frac{2 x}{1 + x ^{2} + e ^{y} + z}

y = sec (x), (\frac{π}{6}, \frac{2 3}{3})

2 x y^{^{'}} - 8 y = x^{- 4}

- \frac{1}{2 1 - x}

\int 1 + t^{4} d t

t \to 0 lim (e^{- 6 t})

What is (\partial)/(\partial x)(ln(1+x^2+e^y+z)) ?
The answer to (\partial)/(\partial x)(ln(1+x^2+e^y+z)) is (2x)/(1+x^2+e^y+z)