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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Treat

y, z

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (1) = 0

\frac{\partial}{\partial x} (x) = 1

= \frac{1}{1 + y} z^{2} (0 + 1)

Simplify

\frac{1}{1 + y} z^{2} (0 + 1) : \frac{z ^{2}}{1 + y}

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

y^{^{''}} - 2 y^{^{'}} - 35 y = 0

\int \frac{x ^{2}}{16 + x ^{6}} d x

\int \frac{( 1 + x ^{2} ) ^{\frac{1}{2}}}{x} d x

x \to + π lim (x cos (\frac{7 x}{6}))

\int 2 x yz e^{x^{2}} d x

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