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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{z ^{2}}{( x + z ) ^{2}}

Solution steps

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

Simplify

y + (\frac{1}{\frac{1}{x} + \frac{1}{z}}) : y + \frac{x z}{z + x}

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

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

n \to \infty lim (∣ n ∣)

\int e^{\frac{4}{x}} d x

ln (2 x - x^{2})

y = x, y = \frac{1}{x ^{2}}, x = 2

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