What is (\partial)/(\partial x)(e^x+y+y^4+6) ?

The answer to (\partial)/(\partial x)(e^x+y+y^4+6) is e^x

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(\partial)/(\partial x)(e^x+y+y^4+6)

\frac{\partial}{\partial x} (e^{x} + y + y^{4} + 6)

e^{x}

Solution steps

\frac{\partial}{\partial x} (e^{x} + y + y^{4} + 6)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (e^{x}) + \frac{\partial}{\partial x} (y) + \frac{\partial}{\partial x} (y^{4}) + \frac{\partial}{\partial x} (6)

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

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

\frac{\partial}{\partial x} (y^{4}) = 0

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

= e^{x} + 0 + 0 + 0

Simplify

= e^{x}

y^{^{''}} + \frac{3}{x} y^{^{'}} - \frac{8}{x ^{2}} y = 0

\int_{- 5}^{5} \frac{1}{x} d x

y^{^{''}} + 10 y^{^{'}} + 25 y = t^{- 2} e^{- 5 t}

\frac{\partial}{\partial x} (e^{2 x} cos (3 y))

\frac{d}{d x} (x (4 x - 5)^{3})

What is (\partial)/(\partial x)(e^x+y+y^4+6) ?
The answer to (\partial)/(\partial x)(e^x+y+y^4+6) is e^x