What is (\partial)/(\partial x)(x^4+y^4-4x+1y) ?

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

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(\partial)/(\partial x)(x^4+y^4-4x+1y)

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x + 1 y)

4 x^{3} - 4

Solution steps

\frac{\partial}{\partial x} (x^{4} + y^{4} - 4 x + 1 \cdot y)

Simplify

x^{4} + y^{4} - 4 x + 1 \cdot y : x^{4} + y^{4} - 4 x + y

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

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

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

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

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

= 4 x^{3} + 0 - 4 + 0

Simplify

= 4 x^{3} - 4

f (x) = 5 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 4

\frac{d y}{d x}, y^{3} + y = x

\int \frac{1}{x ^{13}} d x

\int sec (x) \cdot tan (x) d x

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

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