What is (\partial)/(\partial x)(6xsin(8x^2y)) ?

The answer to (\partial)/(\partial x)(6xsin(8x^2y)) is 6(sin(8x^2y)+16x^2ycos(8x^2y))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(6xsin(8x^2y))

\frac{\partial}{\partial x} (6 x sin (8 x^{2} y))

6 (sin (8 x^{2} y) + 16 x^{2} y cos (8 x^{2} y))

Solution steps

\frac{\partial}{\partial x} (6 x sin (8 x^{2} y))

Treat

y

as a constant

Take the constant out:

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

= 6 \frac{\partial}{\partial x} (x sin (8 x^{2} y))

Apply the Product Rule:

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

f = x, g = sin (8 x^{2} y)

= 6 (\frac{\partial}{\partial x} (x) sin (8 x^{2} y) + \frac{\partial}{\partial x} (sin (8 x^{2} y)) x)

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

\frac{\partial}{\partial x} (sin (8 x^{2} y)) = cos (8 x^{2} y) \cdot 16 y x

= 6 (1 \cdot sin (8 x^{2} y) + cos (8 x^{2} y) \cdot 16 y xx)

Simplify

6 (1 \cdot sin (8 x^{2} y) + cos (8 x^{2} y) \cdot 16 y xx) : 6 (sin (8 x^{2} y) + 16 x^{2} y cos (8 x^{2} y))

= 6 (sin (8 x^{2} y) + 16 x^{2} y cos (8 x^{2} y))

- \frac{2 x}{( 1 + x ^{2} ) ^{2}}

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

y = x^{2}, y = x, y = 2

\frac{d}{d x} (\frac{1 ^{2}}{2 ^{1}})

n = 0 \sum \infty (\frac{1}{2})^{2 n}

What is (\partial)/(\partial x)(6xsin(8x^2y)) ?
The answer to (\partial)/(\partial x)(6xsin(8x^2y)) is 6(sin(8x^2y)+16x^2ycos(8x^2y))