What is (\partial)/(\partial y)(sin(3x+4y)) ?

The answer to (\partial)/(\partial y)(sin(3x+4y)) is cos(3x+4y)*4

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(\partial)/(\partial y)(sin(3x+4y))

\frac{\partial}{\partial y} (sin (3 x + 4 y))

cos (3 x + 4 y) \cdot 4

Solution steps

\frac{\partial}{\partial y} (sin (3 x + 4 y))

Treat

x

as a constant

Apply the chain rule:

cos (3 x + 4 y) \frac{\partial}{\partial y} (3 x + 4 y)

= cos (3 x + 4 y) \frac{\partial}{\partial y} (3 x + 4 y)

\frac{\partial}{\partial y} (3 x + 4 y) = 4

= cos (3 x + 4 y) \cdot 4

\int \frac{cos ( t )}{1 + sin ( t )} d t

\frac{\partial}{\partial x} (x^{2} ln (x))

\int \frac{4 x - 2}{( x ^{2} - x + 1 ) ^{7}} d x

f (x) = ln (cos (2 x))

\int \frac{8}{16 - x ^{2}} d x

What is (\partial)/(\partial y)(sin(3x+4y)) ?
The answer to (\partial)/(\partial y)(sin(3x+4y)) is cos(3x+4y)*4