What is (\partial)/(\partial y)(y^2sin(x/y)) ?

The answer to (\partial)/(\partial y)(y^2sin(x/y)) is 2ysin(x/y)-xcos(x/y)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(y^2sin(x/y))

\frac{\partial}{\partial y} (y^{2} sin (\frac{x}{y}))

2 y sin (\frac{x}{y}) - x cos (\frac{x}{y})

Solution steps

\frac{\partial}{\partial y} (y^{2} sin (\frac{x}{y}))

Treat

x

as a constant

Apply the Product Rule:

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

f = y^{2}, g = sin (\frac{x}{y})

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

\frac{\partial}{\partial y} (y^{2}) = 2 y

\frac{\partial}{\partial y} (sin (\frac{x}{y})) = - \frac{x cos ( \frac{x}{y} )}{y ^{2}}

= 2 y sin (\frac{x}{y}) + - \frac{x cos ( \frac{x}{y} )}{y ^{2}} y^{2}

Simplify

2 y sin (\frac{x}{y}) + - \frac{x cos ( \frac{x}{y} )}{y ^{2}} y^{2} : 2 y sin (\frac{x}{y}) - x cos (\frac{x}{y})

= 2 y sin (\frac{x}{y}) - x cos (\frac{x}{y})

y = x^{3}, y = x^{2} - 2 x, [- 1, 1]

\frac{2}{3} x^{3} + x^{2} - 12 x + 12

f (t) = t

y^{^{'}} = e^{5 x + 4 y}

n = 0 \sum \infty \frac{1}{5}

What is (\partial)/(\partial y)(y^2sin(x/y)) ?
The answer to (\partial)/(\partial y)(y^2sin(x/y)) is 2ysin(x/y)-xcos(x/y)