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

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

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

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

4

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 4 - 0 - 0

Simplify

= 4

\frac{85 t}{99 t - 85}

x \to 3 lim (x^{2} + 2 x - 1)

v + x \cdot \frac{d v}{d x} = x^{2} - 1 + 2 v

b cos (x) + x^{2} sin (x)

\frac{d}{d x} (\frac{( x ^{2} - 1 )}{( x ^{2} + x + 1 )})

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