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

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

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

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

4 x^{3}

Solution steps

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

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^{2}) - \frac{\partial}{\partial x} (4)

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

\frac{\partial}{\partial x} (y^{2}) = 0

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

= 4 x^{3} + 0 - 0

Simplify

= 4 x^{3}

\int sec^{5} (2 x) tan^{3} (2 x) d x

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 3

y^{^{''}} - y^{^{'}} - 2 y = e^{2 t}

5 x^{5}

\frac{d}{d t} (\frac{1}{t})

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