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

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

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(\partial)/(\partial x)(3cos(x+y))

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

- 3 sin (x + y)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

- sin (x + y) \frac{\partial}{\partial x} (x + y)

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

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

= 3 (- sin (x + y) \cdot 1)

Simplify

= - 3 sin (x + y)

x \to - 100 lim (3 x^{2})

\frac{d}{d x} (2 x^{2} - 3 x)

x \to 7 lim (\frac{1}{( x - 7 ) ^{2}})

\frac{d}{d y} (\frac{1}{y} e^{x y} - ln (y^{2}))

x \to - \infty lim (2 x - 3)

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