What is (\partial)/(\partial x)(4x^2cos(xy)) ?

The answer to (\partial)/(\partial x)(4x^2cos(xy)) is 4(2xcos(xy)-x^2ysin(xy))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(4x^2cos(xy))

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

4 (2 x cos (x y) - x^{2} y sin (x y))

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the Product Rule:

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

f = x^{2}, g = cos (x y)

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

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

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

= 4 (2 x cos (x y) + (- sin (x y) y) x^{2})

Simplify

= 4 (2 x cos (x y) - x^{2} y sin (x y))

\frac{d}{d x} (\frac{20 x}{( x ^{2} - 4 ) ^{2}})

\int \frac{6}{x ^{2} - 4} d x

y^{^{'}} = b (1 - y) y

247 x - 19 x^{2}

\int {y (2 a x + b y)} d x

What is (\partial)/(\partial x)(4x^2cos(xy)) ?
The answer to (\partial)/(\partial x)(4x^2cos(xy)) is 4(2xcos(xy)-x^2ysin(xy))