What is (\partial)/(\partial x)(7xsin(4x^2y)) ?

The answer to (\partial)/(\partial x)(7xsin(4x^2y)) is 7(sin(4x^2y)+8x^2ycos(4x^2y))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)(7xsin(4x^2y))

\frac{\partial}{\partial x} (7 x sin (4 x^{2} y))

7 (sin (4 x^{2} y) + 8 x^{2} y cos (4 x^{2} y))

Solution steps

\frac{\partial}{\partial x} (7 x sin (4 x^{2} y))

Treat

y

as a constant

Take the constant out:

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

= 7 \frac{\partial}{\partial x} (x sin (4 x^{2} y))

Apply the Product Rule:

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

f = x, g = sin (4 x^{2} y)

= 7 (\frac{\partial}{\partial x} (x) sin (4 x^{2} y) + \frac{\partial}{\partial x} (sin (4 x^{2} y)) x)

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

\frac{\partial}{\partial x} (sin (4 x^{2} y)) = cos (4 x^{2} y) \cdot 8 y x

= 7 (1 \cdot sin (4 x^{2} y) + cos (4 x^{2} y) \cdot 8 y xx)

Simplify

7 (1 \cdot sin (4 x^{2} y) + cos (4 x^{2} y) \cdot 8 y xx) : 7 (sin (4 x^{2} y) + 8 x^{2} y cos (4 x^{2} y))

= 7 (sin (4 x^{2} y) + 8 x^{2} y cos (4 x^{2} y))

\int \frac{( z ^{2} + 1 ) ^{2}}{z} d z

f (x) = x^{3} - 3 x + 1, (2, 3)

\int cos^{3} (5 - x) sin (5 - x) d x

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

y^{^{'}} = x + y, y (0) = 6

What is (\partial)/(\partial x)(7xsin(4x^2y)) ?
The answer to (\partial)/(\partial x)(7xsin(4x^2y)) is 7(sin(4x^2y)+8x^2ycos(4x^2y))