What is the first d/(dP)((V^1)(q^4+P)) ?

The first d/(dP)((V^1)(q^4+P)) is V

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d/(dP)((V^1)(q^4+P))

\frac{d}{d P} ((V^{1}) (q^{4} + P))

V

Solution steps

\frac{d}{d P} ((V^{1}) (q^{4} + P))

Simplify

(V^{1}) (q^{4} + P) : V (q^{4} + P)

= \frac{d}{d P} (V (q^{4} + P))

Treat

V, q

as constants

Take the constant out:

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

= V \frac{d}{d P} (q^{4} + P)

Apply the Sum/Difference Rule:

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

= V (\frac{d}{d P} (q^{4}) + \frac{d P}{d P})

\frac{d}{d P} (q^{4}) = 0

\frac{d P}{d P} = 1

= V (0 + 1)

Simplify

= V

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