What is (\partial)/(\partial x)(e^y+x+x^2+6) ?

The answer to (\partial)/(\partial x)(e^y+x+x^2+6) is 2x+1

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(\partial)/(\partial x)(e^y+x+x^2+6)

\frac{\partial}{\partial x} (e^{y} + x + x^{2} + 6)

2 x + 1

Solution steps

\frac{\partial}{\partial x} (e^{y} + x + x^{2} + 6)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (e^{y}) + \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (6)

\frac{\partial}{\partial x} (e^{y}) = 0

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

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

\frac{\partial}{\partial x} (6) = 0

= 0 + 1 + 2 x + 0

Simplify

= 2 x + 1

f (x) = \frac{x ^{2} + 5}{x - 5}, at x = 3

cos (\frac{x}{2})

x \to 0 + lim ((1 + 6 x)^{\frac{3}{x}})

\frac{\partial}{\partial t} (arctan (x t))

\int cos (12 x) d x

What is (\partial)/(\partial x)(e^y+x+x^2+6) ?
The answer to (\partial)/(\partial x)(e^y+x+x^2+6) is 2x+1