What is (\partial)/(\partial x)(4+x+x^2Y(x)) ?

The answer to (\partial)/(\partial x)(4+x+x^2Y(x)) is x^2(\partial)/(\partial x)({Y}(x))+2x{Y}(x)+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(4+x+x^2Y(x))

\frac{\partial}{\partial x} (4 + x + x^{2} Y (x))

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

Solution steps

\frac{\partial}{\partial x} (4 + x + x^{2} Y (x))

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (4) + \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial x} (x^{2} Y (x))

\frac{\partial}{\partial x} (4) = 0

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

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

= 0 + 1 + 2 x Y (x) + \frac{\partial}{\partial x} (Y (x)) x^{2}

Simplify

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

\frac{d y}{d x} = x + 2

\int 2 x + 9 x^{- 2} d x

y^{^{'}} - \frac{1}{x} y = x^{2} y^{4}

f (x) = 3 x^{2} - 3 x - 5

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

What is (\partial)/(\partial x)(4+x+x^2Y(x)) ?
The answer to (\partial)/(\partial x)(4+x+x^2Y(x)) is x^2(\partial)/(\partial x)({Y}(x))+2x{Y}(x)+1