What is (\partial)/(\partial x)(3y^2sin(6x)) ?

The answer to (\partial)/(\partial x)(3y^2sin(6x)) is 18y^2cos(6x)

(\partial)/(\partial x)(3y^2sin(6x))

\frac{\partial}{\partial x} (3 y^{2} sin (6 x))

18 y^{2} cos (6 x)

Solution steps

\frac{\partial}{\partial x} (3 y^{2} sin (6 x))

Treat

y

as a constant

Take the constant out:

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

= 3 y^{2} \frac{\partial}{\partial x} (sin (6 x))

Apply the chain rule:

cos (6 x) \frac{\partial}{\partial x} (6 x)

= cos (6 x) \frac{\partial}{\partial x} (6 x)

\frac{\partial}{\partial x} (6 x) = 6

= 3 y^{2} cos (6 x) \cdot 6

Simplify

= 18 y^{2} cos (6 x)

x y^{^{'}} - 3 y^{^{'}} - (x - 3) sin (x) + y = 0

f^{^{'}} (x) = 12 x^{2} - 6 x + 1, f (1) = 5

\frac{d}{d x} (- \frac{1}{6} x^{3} - \frac{1}{2} x^{2} + x + 1 + k e^{k})

\int 8 sec^{4} (θ) tan^{2} (θ) d θ

\frac{\partial}{\partial x} (x y + \frac{1}{x} + \frac{8}{y})

What is (\partial)/(\partial x)(3y^2sin(6x)) ?
The answer to (\partial)/(\partial x)(3y^2sin(6x)) is 18y^2cos(6x)