What is (\partial)/(\partial x)(ln(9x^2+2y^2+4)) ?

The answer to (\partial)/(\partial x)(ln(9x^2+2y^2+4)) is (18x)/(9x^2+2y^2+4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(ln(9x^2+2y^2+4))

\frac{\partial}{\partial x} (ln (9 x^{2} + 2 y^{2} + 4))

\frac{18 x}{9 x ^{2} + 2 y ^{2} + 4}

Solution steps

\frac{\partial}{\partial x} (ln (9 x^{2} + 2 y^{2} + 4))

Treat

y

as a constant

Apply the chain rule:

\frac{1}{9 x ^{2} + 2 y ^{2} + 4} \frac{\partial}{\partial x} (9 x^{2} + 2 y^{2} + 4)

= \frac{1}{9 x ^{2} + 2 y ^{2} + 4} \frac{\partial}{\partial x} (9 x^{2} + 2 y^{2} + 4)

\frac{\partial}{\partial x} (9 x^{2} + 2 y^{2} + 4) = 18 x

= \frac{1}{9 x ^{2} + 2 y ^{2} + 4} \cdot 18 x

Simplify

\frac{1}{9 x ^{2} + 2 y ^{2} + 4} \cdot 18 x : \frac{18 x}{9 x ^{2} + 2 y ^{2} + 4}

= \frac{18 x}{9 x ^{2} + 2 y ^{2} + 4}

\frac{\partial}{\partial x} (x + \frac{y}{z})

y^{^{''}} - 8 y^{^{'}} + 17 y = 0

3 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

3 x^{2}, \frac{1}{3} x

\frac{d y}{d x} = (1 + e^{x}) y

What is (\partial)/(\partial x)(ln(9x^2+2y^2+4)) ?
The answer to (\partial)/(\partial x)(ln(9x^2+2y^2+4)) is (18x)/(9x^2+2y^2+4)