What is (\partial)/(\partial x)(A+B*cos(x)sin(y)) ?

The answer to (\partial)/(\partial x)(A+B*cos(x)sin(y)) is -Bsin(x)sin(y)

(\partial)/(\partial x)(A+B*cos(x)sin(y))

\frac{\partial}{\partial x} (A + B \cdot cos (x) sin (y))

- B sin (x) sin (y)

Solution steps

\frac{\partial}{\partial x} (A + B cos (x) sin (y))

Treat

A, B, y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (A) + \frac{\partial}{\partial x} (B cos (x) sin (y))

\frac{\partial}{\partial x} (A) = 0

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

= 0 - B sin (y) sin (x)

Simplify

= - B sin (x) sin (y)

\frac{d x}{d t} = x^{\frac{2}{3}}

in v erse l a pl a ce \frac{1}{4 s ^{2}}

y^{^{'}} - y = 3 t e^{2 t}, y (0) = 1

\frac{\partial}{\partial x} (y e^{(\frac{x}{y})})

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2} - 4 x}{x + 1})

What is (\partial)/(\partial x)(A+B*cos(x)sin(y)) ?
The answer to (\partial)/(\partial x)(A+B*cos(x)sin(y)) is -Bsin(x)sin(y)