What is (\partial)/(\partial x)((7xy^2)/(x^2+y^4)) ?

The answer to (\partial)/(\partial x)((7xy^2)/(x^2+y^4)) is (7y^2(-x^2+y^4))/((x^2+y^4)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)((7xy^2)/(x^2+y^4))

\frac{\partial}{\partial x} (\frac{7 x y ^{2}}{x ^{2} + y ^{4}})

\frac{7 y ^{2} ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{7 x y ^{2}}{x ^{2} + y ^{4}})

Treat

y

as a constant

Take the constant out:

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

= 7 y^{2} \frac{\partial}{\partial x} (\frac{x}{x ^{2} + y ^{4}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 7 y^{2} \frac{\frac{\partial}{\partial x} ( x ) ( x ^{2} + y ^{4} ) - \frac{\partial}{\partial x} ( x ^{2} + y ^{4} ) x}{( x ^{2} + y ^{4} ) ^{2}}

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

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

= 7 y^{2} \frac{1 \cdot ( x ^{2} + y ^{4} ) - 2 xx}{( x ^{2} + y ^{4} ) ^{2}}

Simplify

7 y^{2} \frac{1 \cdot ( x ^{2} + y ^{4} ) - 2 xx}{( x ^{2} + y ^{4} ) ^{2}} : \frac{7 y ^{2} ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

= \frac{7 y ^{2} ( - x ^{2} + y ^{4} )}{( x ^{2} + y ^{4} ) ^{2}}

\int \frac{cos ( x )}{sin ^{2} ( x ) + 2 sin ( x )} d x

\int 2 e^{- \frac{x}{4}} d x

x \to 9 + lim (6)

y^{^{'}} = y - x + 1

\int b^{4 x} d x

What is (\partial)/(\partial x)((7xy^2)/(x^2+y^4)) ?
The answer to (\partial)/(\partial x)((7xy^2)/(x^2+y^4)) is (7y^2(-x^2+y^4))/((x^2+y^4)^2)