What is (\partial)/(\partial y)(ln(2w^3-7y+3xy^2)) ?

The answer to (\partial)/(\partial y)(ln(2w^3-7y+3xy^2)) is (6yx-7)/(2w^3-7y+3y^2x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(ln(2w^3-7y+3xy^2))

\frac{\partial}{\partial y} (ln (2 w^{3} - 7 y + 3 x y^{2}))

\frac{6 y x - 7}{2 w ^{3} - 7 y + 3 y ^{2} x}

Solution steps

\frac{\partial}{\partial y} (ln (2 w^{3} - 7 y + 3 x y^{2}))

Treat

w, x

as constants

Apply the chain rule:

\frac{1}{2 w ^{3} - 7 y + 3 x y ^{2}} \frac{\partial}{\partial y} (2 w^{3} - 7 y + 3 x y^{2})

= \frac{1}{2 w ^{3} - 7 y + 3 x y ^{2}} \frac{\partial}{\partial y} (2 w^{3} - 7 y + 3 x y^{2})

\frac{\partial}{\partial y} (2 w^{3} - 7 y + 3 x y^{2}) = 6 x y - 7

= \frac{1}{2 w ^{3} - 7 y + 3 x y ^{2}} (6 x y - 7)

Simplify

\frac{1}{2 w ^{3} - 7 y + 3 x y ^{2}} (6 x y - 7) : \frac{6 y x - 7}{2 w ^{3} - 7 y + 3 y ^{2} x}

= \frac{6 y x - 7}{2 w ^{3} - 7 y + 3 y ^{2} x}

\int e^{2 θ} sin (2 θ) d θ

(x - 4) (x + 1)

y = \frac{6 x}{1 - tan ( x )}

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

x \to 0 - lim (b x + 8)

What is (\partial)/(\partial y)(ln(2w^3-7y+3xy^2)) ?
The answer to (\partial)/(\partial y)(ln(2w^3-7y+3xy^2)) is (6yx-7)/(2w^3-7y+3y^2x)