What is (\partial)/(\partial x)(x^4+x^2y^2+y^5+x+y) ?

The answer to (\partial)/(\partial x)(x^4+x^2y^2+y^5+x+y) is 4x^3+2xy^2+1

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(\partial)/(\partial x)(x^4+x^2y^2+y^5+x+y)

\frac{\partial}{\partial x} (x^{4} + x^{2} y^{2} + y^{5} + x + y)

4 x^{3} + 2 x y^{2} + 1

Solution steps

\frac{\partial}{\partial x} (x^{4} + x^{2} y^{2} + y^{5} + x + y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{4}) + \frac{\partial}{\partial x} (x^{2} y^{2}) + \frac{\partial}{\partial x} (y^{5}) + \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial x} (y)

\frac{\partial}{\partial x} (x^{4}) = 4 x^{3}

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

\frac{\partial}{\partial x} (y^{5}) = 0

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

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

= 4 x^{3} + 2 y^{2} x + 0 + 1 + 0

Simplify

= 4 x^{3} + 2 x y^{2} + 1

5 + e^{- 3 t} + t e^{- 4 t}

\int \frac{1}{( e ^{- y} + 1 )} d y

\frac{d}{d x} (e^{x^{2}} + 2 e^{x^{2}} x^{2})

\int_{- 4}^{3} (x^{3} + x^{2} - 6 x) - 6 x d x

y = \frac{1}{x ^{5}}

What is (\partial)/(\partial x)(x^4+x^2y^2+y^5+x+y) ?
The answer to (\partial)/(\partial x)(x^4+x^2y^2+y^5+x+y) is 4x^3+2xy^2+1