What is (\partial)/(\partial y)(e^{-xy^2}+y^3x^4) ?

The answer to (\partial)/(\partial y)(e^{-xy^2}+y^3x^4) is -2xe^{-xy^2}y+3x^4y^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(e^{-xy^2}+y^3x^4)

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

- 2 x e^{- x y^{2}} y + 3 x^{4} y^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= - 2 x e^{- x y^{2}} y + 3 x^{4} y^{2}

x y

n = 1 \sum \infty n^{3} + n

(x^{5})^{^{'}}

\int \frac{1}{4 x + 9} d x

\frac{x}{1 + 3 x ^{2}}

What is (\partial)/(\partial y)(e^{-xy^2}+y^3x^4) ?
The answer to (\partial)/(\partial y)(e^{-xy^2}+y^3x^4) is -2xe^{-xy^2}y+3x^4y^2