What is (\partial)/(\partial x)(3(y^2-x^2)ln(x+y)) ?

The answer to (\partial)/(\partial x)(3(y^2-x^2)ln(x+y)) is 3(-2xln(x+y)+y-x)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(3(y^2-x^2)ln(x+y))

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

3 (- 2 x ln (x + y) + y - x)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = y^{2} - x^{2}, g = ln (x + y)

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

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

\frac{\partial}{\partial x} (ln (x + y)) = \frac{1}{x + y}

= 3 ((- 2 x) ln (x + y) + \frac{1}{x + y} (y^{2} - x^{2}))

Simplify

3 ((- 2 x) ln (x + y) + \frac{1}{x + y} (y^{2} - x^{2})) : 3 (- 2 x ln (x + y) + y - x)

= 3 (- 2 x ln (x + y) + y - x)

\int x sin^{3} (x) cos (x) d x

\frac{d}{d t} (cos (2 t))

\frac{d}{d x} ((x + 1) (2 x + 5)^{4})

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

\frac{x ^{2} - 6 x}{x + 1}

What is (\partial)/(\partial x)(3(y^2-x^2)ln(x+y)) ?
The answer to (\partial)/(\partial x)(3(y^2-x^2)ln(x+y)) is 3(-2xln(x+y)+y-x)