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

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

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

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

4 x^{3} - 4 y

Solution steps

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

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 y) + \frac{\partial}{\partial x} (1)

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

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

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

\frac{\partial}{\partial x} (1) = 0

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

Simplify

= 4 x^{3} - 4 y

\frac{\partial}{\partial x} (ln (3 x^{2} + y^{2} + 4))

\frac{d}{d x} (sec (x) (x - tan (x)))

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- 2 x}

x = 6 y^{2}, x = 4 + 2 y^{2}

f (x) = x (x^{6} - 1), at x = 1

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