What is (\partial)/(\partial x)((y/x)*ln(x)) ?

The answer to (\partial)/(\partial x)((y/x)*ln(x)) is (y(-ln(x)+1))/(x^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/x)*ln(x))

\frac{\partial}{\partial x} ((\frac{y}{x}) \cdot ln (x))

\frac{y ( - ln ( x ) + 1 )}{x ^{2}}

Solution steps

\frac{\partial}{\partial x} ((\frac{y}{x}) ln (x))

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{1}{x}, g = ln (x)

= y (\frac{\partial}{\partial x} (\frac{1}{x}) ln (x) + \frac{\partial}{\partial x} (ln (x)) \frac{1}{x})

\frac{\partial}{\partial x} (\frac{1}{x}) = - \frac{1}{x ^{2}}

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

= y ((- \frac{1}{x ^{2}}) ln (x) + \frac{1}{x} \cdot \frac{1}{x})

Simplify

y ((- \frac{1}{x ^{2}}) ln (x) + \frac{1}{x} \cdot \frac{1}{x}) : \frac{y ( - ln ( x ) + 1 )}{x ^{2}}

= \frac{y ( - ln ( x ) + 1 )}{x ^{2}}

x \to 0 lim (\frac{x}{x - 4})

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = - 12

x \to 2 lim (x - 5)

\int_{0}^{2} \frac{1}{( 2 x - 1 ) ^{\frac{2}{3}}} d x

\frac{\partial}{\partial x} (x^{4} + 3 y^{2} - x)

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