What is (\partial)/(\partial y)(4y^2-20) ?

The answer to (\partial)/(\partial y)(4y^2-20) is 8y

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(\partial)/(\partial y)(4y^2-20)

\frac{\partial}{\partial y} (4 y^{2} - 20)

8 y

Solution steps

\frac{\partial}{\partial y} (4 y^{2} - 20)

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (4 y^{2}) - \frac{\partial}{\partial y} (20)

\frac{\partial}{\partial y} (4 y^{2}) = 8 y

\frac{\partial}{\partial y} (20) = 0

= 8 y - 0

Simplify

= 8 y

f (x) = 8 ln (x) - x^{2}

\frac{d y}{d x} = 10 - 2 y

\frac{d}{d x} (sin (x^{8}))

\frac{\partial}{\partial x} (2 k^{4} + x + 64)

\frac{e ^{- s}}{s + 1}

What is (\partial)/(\partial y)(4y^2-20) ?
The answer to (\partial)/(\partial y)(4y^2-20) is 8y