What is (\partial)/(\partial x)((y+cx),(x+k)) ?

The answer to (\partial)/(\partial x)((y+cx),(x+k)) is (,(c(x+k)+1*(y+cx)))

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(\partial)/(\partial x)((y+cx),(x+k))

\frac{\partial}{\partial x} ((y + c x), (x + k))

(, (c (x + k) + 1 \cdot (y + c x)))

Solution steps

\frac{\partial}{\partial x} ((y + c x), (x + k))

Treat

c, k, y

as a constant

Take the constant out:

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

= (, \frac{\partial}{\partial x} ((y + c x) (x + k)))

Apply the Product Rule:

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

f = y + c x, g = x + k

= (, (\frac{\partial}{\partial x} (y + c x) (x + k) + \frac{\partial}{\partial x} (x + k) (y + c x)))

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

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

= (, (c (x + k) + 1 \cdot (y + c x)))

x \to \infty lim (\frac{1}{6 x})

\frac{d}{d x} ((- x^{2} + 9)^{2})

\frac{\partial}{\partial v} (x^{5} v^{2} + 2 y^{3} u - 3)

\int 2^{\frac{x}{2}} d x

x \cdot \frac{d y}{d x} = 2 x e^{x} - y + 6 x^{2}

What is (\partial)/(\partial x)((y+cx),(x+k)) ?
The answer to (\partial)/(\partial x)((y+cx),(x+k)) is (,(c(x+k)+1*(y+cx)))