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

The answer to (\partial)/(\partial x)(2x^2cos(xy)) is 2(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)(2x^2cos(xy))

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

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

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

= 2 \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)

= 2 (\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

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

Simplify

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

f (x) = (x^{2} + 36) (2 x + 21), at x = - 4

\int \frac{cosh ( x )}{sinh ( x )} d x

x \to 0 + lim (\frac{2}{3 x ^{\frac{1}{3}}})

f (x) = - 4 x^{2}

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

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