What is the d/(dp)((q)(4p^2+q^2)) ?

The d/(dp)((q)(4p^2+q^2)) is 8qp

What is the first d/(dp)((q)(4p^2+q^2)) ?

The first d/(dp)((q)(4p^2+q^2)) is 8qp

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\frac{d}{d p} ((q) (4 p^{2} + q^{2}))

8 qp

Solution steps

\frac{d}{d p} ((q) (4 p^{2} + q^{2}))

Treat

q

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= q \frac{d}{d p} (4 p^{2} + q^{2})

Apply the Sum/Difference Rule:

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

= q (\frac{d}{d p} (4 p^{2}) + \frac{d}{d p} (q^{2}))

\frac{d}{d p} (4 p^{2}) = 8 p

\frac{d}{d p} (q^{2}) = 0

= q (8 p + 0)

Simplify

= 8 qp

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What is the d/(dp)((q)(4p^2+q^2)) ?
The d/(dp)((q)(4p^2+q^2)) is 8qp
What is the first d/(dp)((q)(4p^2+q^2)) ?
The first d/(dp)((q)(4p^2+q^2)) is 8qp